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.
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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.
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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.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
1-4-1
PROBLEM 1-4
Statement:
Units:
Convert a mass of 1 000 lbm to (a) lbf, (b) slugs, (c) blobs, (d) kg.
blob :=
lbf 
sec
2
in
Given:
Mass
Solution:
See Mathcad file P0104.
M := 1000 
lb
1. To determine the weight of the given mass, multiply the mass value by the acceleration due to gravity, g.
W := M 
g
W = 1000 lbf
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
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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
in
a := 40
sec
Solution:
1.
2
See Mathcad file P0105.
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:
F := M 
a
a = 3.333s
-2

ft
F = 25.9 lbf
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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 
Given:
M  100  kg
lbf  sec
2
in
Assumptions: The mass is at sea-level and the gravitational acceleration is
g  32.174
ft
sec
Solution:
1.
or
2
g  386.089 
in
sec
or
2
g  9.807 
m
sec
2
See Mathcad file P0106.
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
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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
2
A  12.000 in
2
A  7742 mm
Moment about x-axis
Moment about y-axis
Ix 
Iy 
b h
3
12
h b
4
Ix  16.000 in
6
4
6
4
Ix  6.660  10  mm
3
4
Iy  9.000  in
12
Iy  3.746  10  mm
Radius of gyration about x-axis
Radius of gyration about y-axis
Polar moment of inertia
kx 
ky 
Ix
kx  1.155  in
A
kx  29.329 mm
Iy
ky  0.866  in
A
ky  21.997 mm
Jz  Ix  Iy
4
Jz  25.000 in
7
4
6
4
6
4
Jz  1.041  10  mm
2.
Solid circle, let:
D  3  in
2
Area
A 
π D
4
Ix 
π D
64
Iy 
π D
64
4
Ix  3.976  in
Ix  1.655  10  mm
4
Moment about y-axis
2
A  4560 mm
4
Moment about x-axis
2
A  7.069  in
4
Iy  3.976  in
Iy  1.655  10  mm
Radius of gyration about x-axis
kx 
Ix
A
kx  0.750  in
kx  19.05  mm
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
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P0107.xmcd
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
Radius of gyration about y-axis
1-7-2
Iy
ky 
ky  0.750  in
ky  19.05  mm
A
4
Jz 
Polar moment of inertia
3.
π D
4
Jz  7.952  in
32
6
4
6
4
6
4
6
4
5
4
5
4
Jz  3.310  10  mm
Hollow circle, let:
D  3  in
d  1  in
A 
Area
Moment about x-axis
Ix 

4
2

4
π
 D d
π
64

2
2
A  6.283  in
2
A  4054 mm
 D d

4
4
Ix  3.927  in
Ix  1.635  10  mm
Moment about y-axis
Iy 

64
π
4
 D d

4
4
Iy  3.927  in
Iy  1.635  10  mm
Radius of gyration about x-axis
Radius of gyration about y-axis
Polar moment of inertia
4.
kx 
ky 
Jz 
Ix
kx  0.791  in
A
kx  20.08  mm
Iy
ky  0.791  in
A
ky  20.08  mm

32
π
4
 D d

4
4
Jz  7.854  in
Jz  3.269  10  mm
Solid semicircle, let:
D  3  in
R  0.5 D
R  1.5 in
2
Area
A 
π D
2
A  3.534  in
8
2
A  2280 mm
Moment about x-axis
Ix  0.1098 R
4
4
Ix  0.556  in
Ix  2.314  10  mm
Moment about y-axis
Iy 
π R
8
4
4
Iy  1.988  in
Iy  8.275  10  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,
P0107.xmcd
mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,
Pearson Education, Inc., Upper Saddle River, NJ 07458.
MACHINE DESIGN - An Integrated Approach, 4th Ed.
Radius of gyration about x-axis
Radius of gyration about y-axis
Polar moment of inertia
1-7-3
kx 
ky 
Ix
A
Iy
A
Jz  Ix  Iy
kx  0.397  in
kx  10.073 mm
ky  0.750  in
ky  19.05  mm
4
Jz  2.544  in
6
4
4
4
4
4
Jz  1.059  10  mm
Distances to centroid
5.
a  0.4244 R
a  0.637  in
a  16.17  mm
b  0.5756 R
b  0.863  in
b  21.93  mm
Right triangle, let:
b  2  in
Area
Moment about x-axis
Moment about y-axis
h  1  in
A 
Ix 
Iy 
b h
2
b h
A  645  mm
3
2
4
Ix  0.056  in
36
h b
2
A  1.000  in
Ix  2.312  10  mm
3
36
4
Iy  0.222  in
Iy  9.250  10  mm
Radius of gyration about x-axis
Radius of gyration about y-axis
Polar moment of inertia
kx 
ky 
Ix
A
Iy
A
Jz  Ix  Iy
kx  0.236  in
kx  5.987  mm
ky  0.471  in
ky  11.974 mm
4
Jz  0.278  in
5
Jz  1.156  10  mm
4
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P0107.xmcd
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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 
Solution:
See the page opposite the inside front cover and Mathcad file P0108.
1.
lbf  sec
2
in
a  2  in
Rectangular prism, let:
b  3  in
c  4  in
3
V  a  b  c
Volume
3
γ  0.28 lbf  in
V  24.000 in
V  393290 mm
M 
Mass
Moment about x-axis
Moment about y-axis
Ix 
Iy 
V γ
3
M  0.017  blob
g
M  3.048  kg
2
M a  b

2
2
Ix  0.019  blob in
12
2
M a  c
Ix  2130.4 kg mm

2
2
Iy  0.029  blob in
12
Iy  3277.6 kg mm
Moment about z-axis
Iz 
2
M b  c

2
Radius of gyration about y-axis
Radius of gyration about z-axis
2.Cylinder, let:
r  2  in
Volume
kx 
ky 
kz 
2
Ix
2
kx  1.041  in
M
kx  26.437 mm
Iy
ky  1.291  in
M
ky  32.791 mm
Iz
kz  1.443  in
M
kz  36.662 mm
3
L  3  in
γ  0.30 lbf  in
2
V  π r  L
3
V  37.699 in
V  617778 mm
Mass
2
Iz  0.036  blob in
12
Iz  4097.0 kg mm
Radius of gyration about x-axis
2
M 
V γ
g
3
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, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
1-8-2
2
Moment about x-axis
Moment about y-axis
Moment about z-axis
Ix 
Iy 
Iz 
M r
2
Ix  0.059  blob in
2

2
Ix  6619.4 kg mm

2
M  3 r  L
2
Iy  0.051  blob in
12

2
Iy  5791.9 kg mm

2
M  3 r  L
2
Radius of gyration about y-axis
Radius of gyration about z-axis
3.
Ix
kx 
kx  35.921 mm
Iy
ky  1.323  in
M
ky  33.601 mm
Iz
kz 
2
kx  1.414  in
M
ky 
2
Iz  0.051  blob in
12
Iz  5791.9 kg mm
Radius of gyration about x-axis
2
kz  1.323  in
M
kz  33.601 mm
Hollow cylinder, let:
a  2  in
b  3  in
Volume
3
L  4  in
γ  0.28 lbf  in
2

2
3
V  π b  a  L
V  62.832 in
V  1029630  mm
Mass
Moment about x-axis
Moment about y-axis
Moment about z-axis
Radius of gyration about x-axis
Radius of gyration about y-axis
M 
Ix 
Iy 
Iz 
kx 
ky 
V γ
M  0.046  blob
g
M  7.98 kg
2
M
 a b
2
3

2
2
Ix  0.296  blob in
4
Ix  3.3  10  kg mm
M
12
M
12

2
2

2
 3 a  3 b  L
2
Iy  0.209  blob in
4
Iy  2.4  10  kg mm

2
2
Ix
M
Iy
M

2
 3 a  3 b  L
2
2
2
Iz  0.209  blob in
4
Iz  2.4  10  kg mm
2
kx  2.550  in
kx  64.758 mm
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.
MACHINE DESIGN - An Integrated Approach, 4th Ed.
Radius of gyration about z-axis
4.
1-8-3
Iz
kz 
kz  2.141  in
M
kz  54.378 mm
Right circular cone, let:
r  2  in
3
h  5  in
γ  0.28 lbf  in
2
Volume
V 
π r  h
3
V  20.944 in
3
V  343210 mm
Mass
Moment about x-axis
Moment about y-axis
M 
Ix 
V γ
M  0.015  blob
g
3
M  2.66 kg
2
10
Iy  M 
3
2
M r
Ix  0.018  blob in
12r2  3h2
80
Ix  2059.4 kg mm
2
Iy  0.023  blob in
Iy  2638.5 kg mm
Moment about z-axis
Iz  M 
12r2  3h2
80
Radius of gyration about y-axis
Radius of gyration about z-axis
5.
Ix
kx 
2
kx  27.824 mm
Iy
ky  1.240  in
M
ky  31.495 mm
Iz
kz 
2
kx  1.095  in
M
ky 
2
Iz  0.023  blob in
Iz  2638.5 kg mm
Radius of gyration about x-axis
2
kz  1.240  in
M
kz  31.495 mm
Sphere, let:
r  3  in
Volume
Mass
Moment about x-axis
V 
M 
Ix 
4
3
3
 π r
V  1853333  mm
V γ
5
3
M  0.082  blob
g
2
3
V  113.097  in
M  14.364 kg
2
M r
2
Ix  0.295  blob in
Ix  33362  kg mm
2
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
Moment about y-axis
Iy 
1-8-4
2
5
2
M r
2
Iy  0.295  blob in
Iy  33362  kg mm
Moment about z-axis
Radius of gyration about x-axis
Radius of gyration about y-axis
Radius of gyration about z-axis
Iz 
kx 
ky 
kz 
2
5
2
M r
2
Iz  0.295  blob in
Iz  33362  kg mm
Ix
M
Iy
M
Iz
M
2
2
kx  1.897  in
kx  48.193 mm
ky  1.897  in
ky  48.193 mm
kz  1.897  in
kz  48.193 mm
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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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 ) 
Moment about y-axis
Iy( b h ) 
3
b h
12
3
h b
12
2
2. Solid circle:
Area
A ( D) 
π D
4
4
Moment about x-axis
Ix( D) 
π D
64
4
Moment about y-axis
3. Hollow circle: Area
Moment about x-axis
Moment about y-axis
Iy( D) 
π D
64
A ( D d ) 
Ix( D d ) 
Iy( D d ) 
π

2
2

4
4

4
4
 D d
4
π
64
π
64
 D d
 D d



4. Solid semicircle:
2
Area
A ( D) 
π D
8
Moment about x-axis
Ix( R)  0.1098 R
Moment about y-axis
Iy( R) 
π R
4
4
8
5. Right triangle:
Area
Moment about x-axis
Moment about y-axis
A ( b h ) 
Ix( b h ) 
Iy( b h ) 
b h
2
b h
3
36
h b
3
36
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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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 γ) 
Moment about x-axis
Moment about y-axis
Moment about z-axis
2.
Ix( a b c γ) 
Iy( a b c γ) 
Iz( a b c γ) 
V ( a b c)  γ
g
2
2
2
2
2
2
M ( a b c γ)  a  b
12
M ( a b c γ)  a  c


12
M ( a b c γ)  b  c

12
Cylinder:
2
Volume
V ( r L)  π r  L
Mass
M ( r L γ) 
V ( r L)  γ
g
2
Moment about x-axis
Moment about y-axis
Moment about z-axis
3.
Ix( r L γ) 
Iy( r L γ) 
Iz( r L γ) 
M ( r L γ)  r
2

2
2
2
12


2
M ( r L γ)  3  r  L
M ( r L γ)  3  r  L

12
Hollow cylinder:
2

2
Volume
V ( a b L)  π b  a  L
Mass
M ( a b L γ) 
Moment about x-axis
Moment about y-axis
Moment about z-axis
Ix( a b L γ) 
Iy( a b L γ) 
Iz( a b L γ) 
V ( a b L)  γ
g
M ( a b L γ)
2
M ( a b L γ)
12
M ( a b L γ)
12
2
 a b

2

2
2
2

2
2
2
 3 a  3 b  L
 3 a  3 b  L


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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4. Right circular cone:
1-10-2
2
Volume
Mass
Moment about x-axis
Moment about y-axis
Moment about z-axis
5.
π r  h
V ( r h ) 
3
M ( r h γ) 
Ix( r h γ) 
V ( r h )  γ
g
3
10
2
 M ( r h γ)  r
2
2

12 r  3  h 
I ( r h γ)  M ( r h γ) 
y
80
Iz( r h γ)  M ( r h γ) 
12r2  3h2
80
Sphere:
Volume
Mass
Moment about x-axis
Moment about y-axis
Moment about z-axis
V ( r) 
4
3
3
 π r
M ( r γ) 
V ( r)  γ
g
Ix( r γ) 
2
Iy( r γ) 
2
Iz( r γ) 
2
5
5
5
2
 M ( r γ)  r
2
 M ( r γ)  r
2
 M ( r γ)  r
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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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)
σya := 5
εya := 5
P2-1(b)
σyb := 9
εyb := 2
P2-1(c)
σyc := 5
εyc := 1
Using equation (2.7), the modulus of resiliency for each material is, approximately,
P21a :=
P21b :=
P21c :=
1
2
1
2
1
2
⋅ σya ⋅ ε ya
P21a = 12.5
⋅ σyb ⋅ ε yb
P21b = 9
⋅ σyc⋅ ε 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.
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
P0201.xmcd
mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,
Pearson Education, Inc., Upper Saddle River, NJ 07458.
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.
2.
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)
S ya := 5
S ua := 6
Figure P2-1(b)
S yb := 9
S ub := 10
Figure P2-1(c)
S yc := 5
S uc := 8
The ratio of yield strength to ultimate strength for each material is:
Figure P2-1(a)
Figure P2-1(b)
Figure P2-1(c)
ratioa :=
ratiob :=
ratioc :=
S ya
S ua
S yb
S ub
S yc
S uc
ratioa = 0.83
ratiob = 0.90
ratioc = 0.63
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P0202.xmcd
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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
Solution:
See Figure 2-19 and Mathcad file P0203.
1.
E  207  GPa
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
Sut 1020  400  MPa
εf 1020  0.365
AISI 1095:
Sy1095  550  MPa
Sut 1095  1050 MPa
εf 1095  0.11
AISI 4142:
Sy4142  1600 MPa
Sut 4142  2430 MPa
εf 4142  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.
2
UR1020 
1 Sy1020

2
E
UR1020  0.22
3
m
2
1 Sy1095
UR1095  
2
E
UR1095  0.73
MN  m
3
m
2
UR4142 
MN  m
1 Sy4142

2
E
UR4142  6.18
MN  m
3
m
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 
UT1095 
UT4142 
1
2
1
2
1
2
  Sy1020  Sut 1020  εf 1020
UT1020  128 
MN  m
3
m
  Sy1095  Sut 1095  εf 1095
UT1095  88
MN  m
3
m
  Sy4142  Sut 4142  εf 4142
UT4142  121 
MN  m
3
m
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.
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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
Solution:
See Figure 2-21 and Mathcad file P0204.
1.
E  71.7 GPa
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
Sut 1100  130  MPa
εf 1100  0.170
2024-T351:
Sy2024  330  MPa
Sut 2024  480  MPa
εf 2024  0.195
7075-T6:
Sy7075  510  MPa
Sut 7075  560  MPa
εf 7075  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.
2
UR1100 
1 Sy1100

2
E
UR1100  0.10
1 Sy2024

2
E
3
m
2
UR2024 
MN  m
UR2024  0.76
MN  m
3
m
2
1 Sy7075
UR7075  
2
E
UR7075  1.81
MN  m
3
m
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 
UT2024 
UT7075 
1
2
1
2
1
2
  Sy1100  Sut 1100  εf 1100
UT1100  21
MN  m
3
m
  Sy2024  Sut 2024  εf 2024
UT2024  79
MN  m
3
m
  Sy7075  Sut 7075  εf 7075
UT7075  88
MN  m
3
m
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 written permission should be
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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
Sut Nylon  80 MPa
εf Nylon  0.52
ENylon  1.1 GPa
HDPE:
SyHDPE  15 MPa
Sut HDPE  23 MPa
εf HDPE  3.0
EHDPE  0.7 GPa
PTFE:
SyPTFE  8.3 MPa
Sut PTFE  13 MPa
εf PTFE  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.
2
URNylon 
1 SyNylon

2 ENylon
URNylon  1.8
MN  m
3
m
2
URHDPE 
1 SyHDPE

2 EHDPE
1 SyPTFE

2 EPTFE
3
m
2
URPTFE 
MN  m
URHDPE  0.16
URPTFE  0.04
MN  m
3
m
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 
UTHDPE 
UTPTFE 
  SyNylon  Sut Nylon  εf Nylon
1
2
2
2
MN  m
3
m
1
1
UTNylon  37
  SyHDPE  SutHDPE  εf HDPE
  SyPTFE  SutPTFE  εf PTFE
UTHDPE  57
MN  m
3
m
UTPTFE  5 
MN  m
3
m
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..
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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
S el  414  MPa
Strain
Test specimen: Diameter d o  12.8 mm
Solution:
1.
Length Lo  50 mm
See Mathcad file P0206.
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 
2.
ε el  0.002
S el
E  207  GPa
ε el
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
 S el ε el
U'el  414 
kN  m
3
m
The total strain energy in the specimen is the strain energy per unit volume times the volume,
Uel  U'el
3.
π d o
4
2
 Lo
Uel  2.7 N  m
Based on the modulus of elasticity and using Table C-1, the material is steel.
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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
S el  41.2 ksi
Strain
Test specimen: Diameter d o  0.505  in
Solution:
1.
S el  284  MPa
Length Lo  2.00 in
See Mathcad file P0207.
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 
2.
ε el  0.004
S el
6
E  10.3 10  psi
ε el
E  71 GPa
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
 S el ε el
U'el  82.4
lbf  in
3
U'el  568 
in
kN  m
3
m
The total strain energy in the specimen is the strain energy per unit volume times the volume,
Uel  U'el
3.
π d o
4
2
 Lo
Uel  33.0 in lbf
Based on the modulus of elasticity and using Table C-1, the material is aluminum.
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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
S el  134  MPa
Strain
Test specimen: Diameter d o  12.8 mm
Solution:
1.
Length Lo  50 mm
See Mathcad file P0208.
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 
2.
ε el  0.003
S el
E  45 GPa
ε el
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
 S el ε el
U'el  201 
kN  m
3
m
The total strain energy in the specimen is the strain energy per unit volume times the volume,
Uel  U'el
3.
π d o
4
2
 Lo
Uel  1.3 N  m
Based on the modulus of elasticity and using Table C-1, the material is magnesium.
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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
S el  100  ksi
Strain
ε el  0.006
S el  689  MPa
Test specimen: Diameter d o  0.505  in
Solution:
1.
See Mathcad file P0209.
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 
2.
Length Lo  2.00 in
S el
6
E  16.7 10  psi
ε el
E  115  GPa
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
 S el ε el
U'el  300 
lbf  in
3
3 kN  m
U'el  2  10 
in
3
m
The total strain energy in the specimen is the strain energy per unit volume times the volume,
Uel  U'el
3.
π d o
4
2
 Lo
Uel  120.18 in lbf
Based on the modulus of elasticity and using Table C-1, the material is titanium.
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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  joule
Given:
Yield strength
S y  689  MPa
Yield strain
ε y  0.006
Solution:
1.
6
See Mathcad file P0210.
The modulus of resilience (strain energy per unit volume) is given by Equation (2.7) and is approximately
UR 
1
2
 S y ε y
UR  2.067 
MJ
3
UR  2.1 MPa
m
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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  joule
Given:
Yield strength
S y  60 ksi
Yield strain
ε y  0.002
Solution:
1.
6
S y  414  MPa
See Mathcad file P0211.
The modulus of resilience (strain energy per unit volume) is given by Equation (2.7) and is approximately
UR 
1
2
 S y ε y
UR  60
in lbf
3
in
UR  0.414 
MJ
3
UR  0.414  MPa
m
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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:
S y  414  MPa
Solution:
See Mathcad file P0212.
1.
ε f  0.15
Determine the modulus of toughness using Equation (2.8).
UT 
2.
S ut  689  MPa
 Sy  S ut 

  εf
 2 
UT  82.7
MN  m
3
UT  82.7 MPa
m
Determine the modulus of resilience using Equation (2.7) and Young's modulus for steel: E  207  GPa
2
1 Sy
UR  
2 E
UR  414 
kN  m
3
UR  0.41 MPa
m
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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
Solution:
See Mathcad file P0213.
HB  250
1. Determine the approximate tensile strength of the material from equations (2.10), not Table 2-3.
S ut  0.5 HB ksi
S ut  125  ksi
S ut  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
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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
Solution:
See Mathcad file P0214.
HB  340
1. Determine the approximate tensile strength of the material from equations (2.10), not Table 2-3.
S ut  0.5 HB ksi
S ut  170  ksi
S ut  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
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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).
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P0215.xmcd
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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).
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P0216.xmcd
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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).
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P0217.xmcd
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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.
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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
Steel
st  0
Ultimate Strength
Sut  80 ksi
st
Young's Modulus
6
E  30 10  psi
st
Weight Density
lbf
γ  0.28
st
3
in
Aluminum
al  1
Sut
al
 60 ksi
E
6
al
 10.4 10  psi
γ  0.10
lbf
al
3
in
Titanium
ti  2
Sut  90 ksi
ti
6
E  16.5 10  psi
ti
γ  0.16
ti
lbf
3
in
Index
Solution:
1.
i  0 1  2
See Mathcad file P0219.
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.
Sut
Specific strength
γ
i 1

i
in
E

286·103
600·103
563·103
2.
Specific stiffness
i 1

γ in

i
Steel
Aluminum
Titanium
107·106
104·106
103·106
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.
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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).
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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
Mat  "2024 Aluminum, HT" Sy  290  MPa
1
1
Specific Weight
3
γ  0.10 lbf  in
γ  27.14 
1
1
kN
3
m
Mat  "1040 CR Steel"
2
Sy  490  MPa
2
3
γ  0.28 lbf  in
γ  76.01 
2
2
kN
3
m
Mat  "Ti-75A Titanium"
3
Sy  517  MPa
3
3
γ  0.16 lbf  in
γ  43.43 
3
3
kN
3
m
Mat  "Type 302 CR SS"
4
Sy  1138 MPa
4
3
γ  0.28 lbf  in
γ  76.01 
4
4
kN
3
m
i  1 2  4
Solution:
1.
See Mathcad file P0221.
Calculate the strength-to-weight ratio for each material as described in Section 2.1.
SWR 
i
Sy
SWR
γ
10  m
i
i
4
i
 "2024 Aluminum, HT" 
"1040 CR Steel"

Mat  
i  "Ti-75A Titanium" 
 "Type 302 CR SS" 



1.068
0.645
1.190
1.497
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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
3
Mat  "2024 Aluminum, HT" Sut  441  MPa
γ  0.10 lbf  in
Mat  "1040 CR Steel"
Sut  586  MPa
γ  0.28 lbf  in
Mat  "Acetal, unfilled"
Sut  60.7 MPa
γ  0.051  lbf  in
Mat  "Ti-75A Titanium"
Sut  586  MPa
γ  0.16 lbf  in
Mat  "Type 302 CR SS"
Sut  1310 MPa
γ  0.28 lbf  in
1
1
2
3
4
5
2
3
4
5
3
γ  27.14  kN  m
1
1
3
3
γ  76.01  kN  m
2
2
3
3
3
3
3
γ  43.43  kN  m
4
4
3
5
3
γ  13.84  kN  m
3
γ  76.01  kN  m
5
i  1 2  5
Solution:
1.
See Mathcad file P0222.
Calculate the strength-to-weight ratio for each material as described in Section 2.1.
Sut
SWR 
i
γ
SWR
i
i
4

10  m
i
 "2024 Aluminum, HT" 
 "1040 CR Steel" 


Mat   "Acetal, unfilled" 
i
 "Ti-75A Titanium" 


 "Type 302 CR SS" 
1.625
0.771
0.438
1.349
1.724
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
PROBLEM 2-23
2-23-1
_____
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  kg
Given:
Material
3
Modulus of Elasticity
Density
3
Mat  "Aluminum"
E  71.7 GPa
ρ  2.8 Mg m
Mat  "Titanium"
E  113.8  GPa
ρ  4.4 Mg m
Mat  "Gray cast iron"
E  103.4  GPa
ρ  7.2 Mg m
Mat  "Ductile iron"
E  168.9  GPa
ρ  6.9 Mg m
Mat  "Bronze"
E  110.3  GPa
ρ  8.6 Mg m
Mat  "Carbon steel"
E  206.8  GPa
ρ  7.8 Mg m
Mat  "Stainless steel"
E  189.6  GPa
ρ  7.8 Mg m
1
1
2
2
3
3
4
4
5
5
6
6
7
7
1
3
2
3
3
3
4
3
5
3
6
3
7
i  1 2  7
Solution:
1.
See Mathcad file P0223.
Calculate the specific stiffness for each material as described in Section 2.1.
E
E' 
i
2.
ρ
E'
i
i
 "Aluminum" 


 "Titanium" 
 "Gray cast iron" 
Mat   "Ductile iron" 
i 

 "Bronze" 
 "Carbon steel" 
 "Stainless steel" 


10
i
6
2

s
2

m
25.6
25.9
14.4
24.5
12.8
26.5
24.3
Rank them in increasing order of specific stiffness.
E'
Mat  "Bronze"
2
5 s
5
10
E'
Mat  "Gray cast iron"
10
E'

2
6

2
10
6
 14.4
m
2
7 s
7
 12.8
m
2
3 s
3
Mat  "Stainless steel"
6

2
 24.3
m
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
E'
Mat  "Ductile iron"
2
4 s
4
10
E'
Mat  "Aluminum"
10
E'
10
E'
6

2
6
10
6
 24.5
 25.6
m

2
 25.9
m
2
6 s
6
3.
2
m
2
2 s
2
Mat  "Carbon steel"

2
1 s
1
Mat  "Titanium"
6
2-23-2

2
 26.5
m
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.
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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.
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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.
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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.
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P0226.xmcd
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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
Solution:
See Mathcad file P0227.
1.
S y  117  ksi
S y  807  MPa
From Table 2-2 the reliability factor for 99.9% is Re  0.753. Applying this to the mean tensile strength gives
S y99.9  S y Re
S y99.9  88.1 ksi
S y99.9  607  MPa
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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
Solution:
See Mathcad file P0228.
1.
S ut  213  ksi
S ut  1469 MPa
From Table 2-2 the reliability factor for 99.99% is Re  0.702. Applying this to the mean ultimate tensile
strength gives
S ut99.99  S ut Re
S ut99.99  150  ksi
S ut99.99  1031 MPa
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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
Solution:
See Mathcad file P0229.
1.
S ut  236  ksi
S ut  1627 MPa
From Table 2-2 the reliability factor for 90% is Re  0.897. Applying this to the mean ultimate
tensile strength gives
S ut99.99  S ut Re
S ut99.99  212  ksi
S ut99.99  1460 MPa
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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
Solution:
See Mathcad file P0230.
S y  165  ksi
S y  1138 MPa
1. From Table 2-2 the reliability factor for 99.999% is Re  0.659. Applying this to the mean
tensile
strength gives
S y99.9  S y Re
S y99.9  109  ksi
S y99.9  750  MPa
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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.
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P0231.xmcd
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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.
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P0232.xmcd
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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.
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P0233.xmcd
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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.
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P0234.xmcd
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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
Solution:
See Mathcad file P0235.
1.
Use equation (2.10), solving for the Brinell hardness, HB.
S ut = 3.45 HB
2.
S ut  550  MPa
HB 
S ut
3.45 MPa
HB  159
From Table 2-3, the equivalent hardness on the Rocwell scale is 83.9HRB.
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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
Solution:
See Mathcad file P0236.
1.
HB  220
Use equation (2.10), solving for ultimate tensile strength.
Minimum:
S utmin  ( 3.45 HB  0.2 HB)  MPa
S utmin  715  MPa
Maximum:
S utmax  ( 3.45 HB  0.2 HB)  MPa
S utmax  803  MPa
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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 f2/3/, 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?
3
Units:
Mg  kg
Given:
5052 Aluminum
3
S ya  255  MPa
ρa  2.8 Mg m
3
CA-170 beryllium copper S yb  1172 MPa
4130 steel
Solution:
ρb  8.3 Mg m
3
S ys  703  MPa
ρs  7.8 Mg m
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 S y ρ  
3
0.667
Sy
ρ

Mg m
0.667
MPa
Aluminum
Ia  Index S ya ρa 
Ia  14.4
Beryllium copper
Ib  Index S yb ρb 
Ib  13.4
Steel
Is  Index S ys ρs 
Is  10.2
The 5052 aluminum has the highest value of the index and would be the best choice to minimize weight.
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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Pearson Education, Inc., Upper Saddle River, NJ 07458.
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?
3
Units:
Mg  kg
Given:
5052 Aluminum
3
S ya  255  MPa
ρa  2.8 Mg m
CA-170 beryllium copper S yb  1172 MPa
4130 steel
Solution:
3
ρb  8.3 Mg m
3
S ys  703  MPa
ρs  7.8 Mg m
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 S y ρ  
S y Mg m 3

MPa
ρ
Aluminum
Ia  Index S ya ρa 
Ia  91.1
Beryllium copper
Ib  Index S yb ρb 
Ib  141.2
Steel
Is  Index S ys ρs 
Is  90.1
The beryllium copper has the highest value of the index and would be the best choice to minimize weight.
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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?
3
Units:
Mg  kg
Given:
5052 Aluminum
ρa  2.8 Mg m
3
CA-170 beryllium copper Eb  127.6  GPa
ρb  8.3 Mg m
Es  206.8  GPa
ρs  7.8 Mg m
4130 steel
Solution:
3
Ea  71.7 GPa
3
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 ρ ) 
E
3
0.5
ρ

Mg m
GPa
0.5
Aluminum
Ia  Index Ea ρa 
Ia  3.0
Beryllium copper
Ib  Index Eb ρb 
Ib  1.4
Steel
Is  Index Es ρs 
Is  1.8
The 5052 aluminum has the highest value of the index and would be the best choice to minimize weight.
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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
Given:
5052 Aluminum
3
ρa  2.8 Mg m
3
CA-170 beryllium copper Eb  127.6  GPa
ρb  8.3 Mg m
Es  206.8  GPa
ρs  7.8 Mg m
4130 steel
Solution:
3
Ea  71.7 GPa
3
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.
3
Index( E ρ ) 
E Mg m

ρ GPa
Aluminum
Ia  Index Ea ρa 
Ia  25.6
Beryllium copper
Ib  Index Eb ρb 
Ib  15.4
Steel
Is  Index Es ρs 
Is  26.5
The steel has the highest value of the index and would be the best choice to minimize weight.
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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)
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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:
1.
See Mathcad file P0302a.
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.
Ma
Rb
Ra
A
B
Fbr
Fax
Fct
Fcr
Rc
α
C
Fay
Rd
Re
Fey
Fex
E
Fdx
D
Fdy
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.
Σ Fx :
−Fax − Fbr⋅ cos( α) + Fcr⋅ cos( α) − Fct⋅ sin( α) − Fdx + Fex = 0
(1)
Σ Fy :
−Fay − Fbr⋅ sin( α) + Fcr⋅ sin( α) + Fct⋅ cos( α) − Fdy + Fey = 0
(2)
Σ Mz:
Ma + ( Rbx ⋅ Fby − Rby⋅ Fbx ) + ( Rcx⋅ Fcy − Rcy⋅ Fcx) ... = 0
+ ( R ⋅ F − R ⋅ F ) + ( R ⋅ F − R ⋅ F )

ex ey
ey ex
 dx dy dy dx

(3)
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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.
Fhand
Tpipe
Fhand
a
(a) FBD of pipe wrench and pipe
Fbt
Fbn
Fax
A
α
Fay
d
b
(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.
4.
Applying equations 3.3b to the FBD of the pipe wrench,
Σ Fx :
−Fax + Fbn⋅ cos( α) − Fbt⋅ sin( α) = 0
(1)
Σ Fy :
−Fay + Fbn⋅ sin( α) + Fbt⋅ cos( α) − Fhand = 0
(2)
Σ M A:
d ⋅ ( Fbt⋅ cos( α) + Fbn⋅ sin( α) ) − ( d + a ) ⋅ Fhand = 0
(3)
These equations can be solved for the vertical forces if we assume α is small so that sin(α) = 0 and cos (α) = 1.
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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.
z
Fchain
Sprocket
Faz
Fbz
a
Frider
A
B
Arm
b
Arm (sectioned)
Fax
Fbx
y
Pedal
x
(a) FBD of complete pedal-arm assembly
z
a
Tc
Frider
b
Mc
Arm
Fc
y
Pedal
x
(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
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
 Ty-axis:
3-3-2
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
Mc  a  Frider
Mc  255  N  m
Tc  b  Frider
Tc  90 N  m
Bending moment in pedal arm at section C
 My-axis:
a  Frider  Mc = 0
Twisting moment in pedal arm at section C
 Mx-axis:
b  Frider  Tc = 0
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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
Mtongue  100  kg
b  31 mm
Fpull  4.905  kN
c  70 mm
t  19 mm
d  20 mm
Assumptions: 1. The nuts are just snug-tight (no pre-load), which is the worst case.
2. All reactions will be concentrated loads rather than distributed loads or pressures.
Solution:
1.
See Figure 3-4 and Mathcad file P0304.
The weight on the tongue is
Wtongue  Mtongue g
Wtongue  0.981  kN
2. The FBD of the hitch and bracket assembly is shown in Figure 3-4. The known external forces that act on the
ball are Fpull and Wtongue . The reactions on the bracket are at points C and D. The bolts at C provide tensile
(Fc2x) and shear (Fc2y) forces, and the bracket resists rotation about point D where the reaction force Fd2 is
applied by the channel to which the bracket is bolted.
3. Solving for the reactions by summing the horizontal and vertical forces and the moments about D:
 Fx :
 Fpull  Fc2x  Fd2 = 0
(1)
 Fy :
Fc2y  Wtongue = 0
(2)
 MD :
Fc2x d  Fpull  ( a  t  b  d )  Wtongue c = 0
(3)
W tongue
70 = c
1
F pull
1
40 = a
2
B
A
2
19 = t
31 = b
C
Fc2x
20 = d
D
B
C
D
Fd2
F c2y
FIGURE 3-4
Dimensions and Free Body Diagram for Problem 3-4
4.
Solving equation (3) for Fc2x
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MACHINE DESIGN - An Integrated Approach,4th Ed.
Fc2x 
5.
Fpull  ( a  t  b  d )  Wtongue c
d
(4)
Fd2  25.505 kN
(5)
Fc2y  0.981  kN
(6)
Solving (2) for Fc2y
Fc2y  Wtongue
7.
Fc2x  30.41  kN
Substituting into (1) and solving for Fd2
Fd2  Fc2x  Fpull
6.
3-4-2
The loads applied to the two bolts that attach the bracket to the channel are:
Axial force on two bolts
Fc2x  30.4 kN
Shear force on two bolts
Fc2y  0.98 kN
We assume that each bolt would carry one half of these loads.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-5-1
PROBLEM 3-5
Statement:
For the trailer hitch of Problem 3-4, determine the horizontal force that will result on the ball from
accelerating a 2000-kg trailer to 60 m/sec in 20 sec.
Given:
Mass of trailer
Mtrailer  2000 kg
Final velocity
vf  60
Time to reach velocity
τ  20 sec
m
sec
Assumptions: 1. Acceleration is constant.
2. The rolling resistance of the tires and the wheel bearings is negligible.
Solution:
1.
See Mathcad file P0305.
From elementary kinematics, the acceleration required is
a 
2.
vf
τ
a  3.00
m
sec
(1)
2
Using Newton's second law to find the force required to accelerate the trailer,
Fhitch  Mtrailer  a
Fhitch  6.00 kN
(2)
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-6-1
PROBLEM 3-6
Statement:
For the trailer hitch of Problem 3-4, determine the horizontal force that will result on the ball from
an impact between the ball and the tongue of the 2000-kg trailer if the hitch deflects 2.8 mm
dynamically on impact. The tractor weighs 1000 kg. The velocity at impact is 0.3 m/sec.
Given:
Mass of trailer
Mtrailer  2000 kg
Dynamic deflection
δi  2.8 mm
Mass of tractor
Mtractor  1000 kg
Impact velocity
vi  0.3
m
sec
Assumptions: 1. The tractor is the "struck member" because the hitch is on the tractor and it is the hitch that
deflects.
2. Equations (3.9) and (3.10) can be used to model the impact.
Solution:
1.
See Mathcad file P0306.
The weight of the trailer (the "striking member") is
Wtrailer  Mtrailer  g
2.
The correction factor, from equation (3-15), is
1
η 
1
3.
Wtrailer  19.613 kN
Mtractor
η  0.857
3  Mtrailer
Eliminating E from equations (3.9a) and (3.10) and solving for the horizontal force on the ball Fi yields
1
2
 Fi δi = η   Mtrailer  vi 
2
2

1
Fi 
η Mtrailer  vi
δi
2
Fi  55.1 kN
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-7-1
PROBLEM 3-7
Statement:
The piston of an internal-combustion engine is connected to its connecting rod with a "wrist
pin." Find the force on the wrist pin if the 0.5-kg piston has an acceleration of 2 500 g.
Given:
Mass of piston
Mpiston  0.5 kg
Acceleration of piston
a piston  2500 g
Assumptions: The force on the wrist pin due to the weight of the piston is very small compared with the
acceleration force.
Solution:
1.
See Mathcad file P0307.
The acceleration in m/s is
4
a piston  2.452  10 
m
sec
2.
2
Using Newton's Second Law expressed in equation (3.1a), the force on the wrist pin is
Fwristpin  Mpiston  a piston
Fwristpin  12.258 kN
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
PROBLEM 3-8
3-8-1
_____
Statement:
A cam-follower system similar to that shown in Figure 3-15 has a mass m = 1 kg, a spring
constant k = 1000 N/m, and a damping coefficient d = 19.4 N-s/m. Find the undamped and
damped natural frequencies of this system.
Units:
cps := 2 ⋅ π⋅ rad⋅ sec
Given:
Mass
−1
M := 1 ⋅ kg,
Spring constant
−1
k := 1000⋅ N ⋅ m
−1
Damping coefficient d := 19.4⋅ N ⋅ s⋅ m
Solution:
1.
See Figure 3-15 and Mathcad file P0308.
Calculate the undamped natural frequency using equation 3.4.
ωn :=
2.
k
ωn = 31.6
M
rad
sec
ωn = 5.03 cps
Calculate the undamped natural frequency using equation 3.7.
ωd :=
d 
− 

M  2⋅ M 
k
2
ωd = 30.1
rad
sec
ωd = 4.79 cps
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-9-1
PROBLEM 3-9
Statement:
A ViseGrip plier-wrench is drawn to scale in Figure P3-3. Scale the drawing for dimensions. Find
the forces acting on each pin and member of the assembly for an assumed clamping force of P =
4000 N in the position shown. What force F is required to keep it in the clamped position shown?
Given:
Clamping force
Dimensions
P  4.00 kN
a  50.0 mm
b  55.0 mm
c  39.5 mm
d  22.0 mm
e  28.0 mm
f  26.9 mm
g  2.8 mm
h  21.2 mm
α  21.0 deg
β  129.2  deg
Assumptions: Links 3 and 4 are in a toggle position, i.e., the pin that joins links 3 and 4 is in line with the pins
that join 1 with 4 and 2 with 3.
Solution:
1.
See Figure 3-9 and Mathcad file P0309.
The FBDs of the assembly and each individual link are shown in Figure 3-9. The dimensions, as scaled from
Figure P3-3 in the text, are given above and are shown on the link FBDs.
4
F
P
1
2
3
P
F
55.0 = b
50.0 = a
39.5 = c
F
F14
22.0 = d
129.2°
1

4
F34
F41
F21
P

28.0 = e


F43

F12
3
F23
F32
P
2.8 = g
21.2 = h
2
F
26.9 = f
FIGURE 3-9
Free Body Diagrams for Problem 3-9
2.
Looking first at Part 3, we see that it is a three-force body. Therefore, the lines of action of the three forces
must intersect at a point. But, since Parts 3 and 4 are in a toggle position, F43 and F23 are colinear, which
means that their x- and y-components must be equal and opposite, leading to the conclussion that F = 0.
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P0309.xmcd
3-9-1
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3.
3-9-2
Now, looking at Part 1, we see that (for F = 0) it is also a three-force body, as is Part 2. In fact, the forces on
Part 1 are identical to those on Part 2. Solving for the unknown reactions on Parts 1 and 2,
 Fx:
F41 cos( 180  deg  α)  F21 cos( β  180  deg) = 0
(a)
 Fy:
F41 sin( 180  deg  α)  F21 sin( β  180  deg)  P = 0
(b)
Solving equation (a) for F21
F21 = 
F41 cos( 180  deg  α)
(c)
cos( β  180  deg)
Substituting equation (c) into (b)
F41 sin( 180  deg  α) 
F41 cos( 180  deg  α)
cos( β  180  deg)
 sin( β  180  deg)  P = 0
(d)
Solving equation (d) for F41
P
F41  
sin( 180  deg  α) 
F21  
cos( 180  deg  α)
cos( β  180  deg)
 sin( β  180  deg)
F41 cos( 180  deg  α)
F41  5.1 kN
cos( β  180  deg)
F21  7.5 kN
Checking moment balance on Part 1,
F41 sin( α)  c  F21 sin( β  90 deg)  d  P a  0  kN  m
The result is, within the accuracy of the scaled dimensions, zero as it must be.
4.
5.
6.
The x and y components of the pin forces on Part 1 are
F41x  F41 cos( 180  deg  α)
F41x  4.749  kN
F41y  F41 sin( 180  deg  α)
F41y  1.823  kN
F21x  F21 cos( β  180  deg)
F21x  4.749  kN
F21y  F21 sin( β  180  deg)
F21y  5.823  kN
The forces on the pins at the ends of Part 4 are
F14  F41
F14  5.1 kN
F34  F14
F34  5.1 kN
The forces on the pins at the ends of Part 3 are
F43  F34
F43  5.1 kN
F23  F43
F23  5.1 kN
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
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P0309.xmcd
3-9-2
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
7.
3-9-3
The forces on the pins at the ends of Part 2 are
F12  F21
F12  7.5 kN
F32  F23
F32  5.1 kN
Checking moment equilibrium on Part 2,
F12 ( e cos( β  90 deg)  g  sin( β  90 deg) )   0  kN  m
 F32 ( h  cos( α)  f  sin( α) )
which is zero, as it must be.
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P0309.xmcd
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-10-1
PROBLEM 3-10
Statement:
Given:
An overhung diving board is shown in Figure P3-4a. Find the reaction forces and construct the
shear and moment diagrams for this board with a 100 kg person standing at the free end.
Determine the maximum shear force, maximum moment and their locations.
Beam length
Distance to support
Mass at free end
L  2000 mm
a  700  mm
M  100  kg
2000 = L
R1
Assumptions: The weight of the beam is negligible
compared to the applied load and so can
be ignored.
P
R2
700 = a
Solution:
See Figure 3-10 and Mathcad file P0310.
FIGURE 3-10A
Free Body Diagram for Problem 3-10
1. From inspection of Figure 3-10, write the load function equation
q(x) = -R1<x - 0>-1 + R2<x - a>-1 - P<x - L >-1
2. Integrate this equation from - to x to obtain shear, V(x)
V(x) = -R1<x - 0>0 + R2<x - a>0 - P<x - L >0
3. Integrate this equation from - to x to obtain moment, M(x)
M(x) = -R1<x - 0>1 + R2<x - a>1 - P<x - L >1
4. Determine the magnitude of the force, P
P  M  g
P  980.7  N
5. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = L, where
both are zero.
At x = L+, V = M = 0
R1  P
La
a
R2  P  R1
6. Define the range for x
V = R1  R2  P = 0
M = R1 L  R2 ( L  a ) = 0
R1  1821 N
R2  2802 N
x  0  in 0.005  L  L
7. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
S ( x z)  if ( x  z 1 0 )
8. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the
effect of the singularity functions.
V ( x)  R1 S ( x 0  in)  R2 S ( x a )  P S ( x L)
M ( x)  R1 S ( x 0  in)  x  R2 S ( x a )  ( x  a )  P S ( x L)  ( x  L)
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-10-2
9. Plot the shear and moment diagrams.
1000
Shear
Diagram
0
V ( x)
N
 1000
 2000
0
500
1000
1500
2000
x
mm
0
Moment
Diagram
 375
M ( x)
Nm
 750
 1125
 1500
0
500
1000
1500
2000
x
mm
FIGURE 3-10B
Shear and Moment Diagrams for Problem 3-10
10. The maximum value of the shear force ocuurs throughout the distance from x = 0 to x = a and is
R1  1821 N
11. Find the maximum value of the bending moment by determining the value of x where the shear is zero.
Inspection of the shear diagram shows that this occurs at x = a.
Mmax  M ( a )
Mmax  1275 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,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-11-1
PROBLEM 3-11
Statement:
Given:
Determine the impact force and dynamic deflection that will result when the 100-kg person in
Problem 3-10 jumps up 250 mm and lands back on the board. Assume that the board weighs
29 kg and deflects 131 mm statically when the person stands on it. Find the reaction forces
and construct the shear and moment diagrams for this dynamic loading. Determine the
maximum shear force, maximum moment and their locations along the length of the board.
Beam length
L  2000 mm
Distance to support
a  700  mm
Mass of person
mpers  100  kg
Mass of board
mboard  29 kg
Static deflection
δst  131  mm
Height of jump
h  250  mm
2000 = L
R1
R2
700 = a
Assumptions: Equation (3.15) can be used to approximate a
mass correction factor.
Solution:
Fi
FIGURE 3-11A
Free Body Diagram for Problem 3-11
See Figure 3-11 and Mathcad file P0311.
1. The person is the moving object and the board is the struck object. The mass ratio to be used in equation
(3.15) for the correction factor is
massratio 
mpers
massratio  3.448
mboard
2. From equation (3.15), the correction factor is
1
η 
1
η  0.912
1
3  massratio
3. The weight of the moving mass is
Wpers  mpers g
Wpers  0.981  kN
4. The dynamic force is found by solving equation (3.14) for Fi.

Fi  Wpers  1 

1
2  η h 
Fi  3.056  kN

δst 
From this we see that the dynamic force ratio is
Fi
Wpers
 3.12
5. From inspection of Figure 3-11, write the load function equation
q(x) = -R1<x - 0>-1 + R2<x - a>-1 - Fi<x - L >-1
6. Integrate this equation from - to x to obtain shear, V(x)
V(x) = -R1<x - 0>0 + R2<x - a>0 - Fi<x - L >0
7. Integrate this equation from - to x to obtain moment, M(x)
M(x) = -R1<x - 0>1 + R2<x - a>1 - Fi<x - L >1
© 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,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-11-2
8. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = L,
where both are zero.
V = R1  R2  Fi = 0
At x = L+, V = M = 0
R1  Fi
L a
M = R1 L  R2 ( L  a ) = 0
R1  5676 N
a
R2  Fi  R1
R2  8733 N
x  0  in 0.005  L  L
9. Define the range for x
10. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than
z, and a value of one when it is greater than or equal to z.
S ( x z)  if ( x  z 1 0 )
11. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get
the effect of the singularity functions.
V ( x)  R1 S ( x 0  in)  R2 S ( x a )  Fi S ( x L)
M ( x)  R1 S ( x 0  in)  x  R2 S ( x a )  ( x  a )  Fi S ( x L)  ( x  L)
12. Plot the shear and moment diagrams.
Shear Diagram
V ( x)
kN
Moment Diagram
4
0
2
1
0
M ( x)
2
kN  m
2
3
4
6
0
0.5
1
1.5
2
4
0
0.5
1
x
x
m
m
1.5
2
FIGURE 3-11B
Shear and Moment Diagrams for Problem 3-11
13. The maximum value of the shear force ocuurs throughout the distance from x = 0 to x = a and is
R1  5676 N
14. Find the maximum value of the bending moment by determining the value of x where the shear is zero.
Inspection of the shear diagram shows that this occurs at x = a.
Mmax  M ( a )
Mmax  3973 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,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-12-1
PROBLEM 3-12
Statement:
Given:
An overhung diving board is shown in Figure P3-4b. Find the reaction forces and construct the
shear and moment diagrams for this board with a 100 kg person standing at the free end.
Determine the maximum shear force, maximum moment and their locations.
Beam length
L  1300 mm
Mass at free end
M  100  kg
Assumptions: 1. The weight of the beam is negligible
compared to the applied load and so can
be ignored.
Solution:
2000
1300 = L
P
M1
R1
See Figure 3-12 and Mathcad file P0312.
700
1. From inspection of Figure 3-12, write the load function
equation
q(x) = -M1<x - 0>-2 + R1<x - a>-1 - P<x - L >-1
FIGURE 3-12A
Free Body Diagram for Problem 3-12
2. Integrate this equation from - to x to obtain shear, V(x)
V(x) = -M1<x - 0>-1 + R1<x - a>0 - P<x - L >0
3. Integrate this equation from - to x to obtain moment, M(x)
M(x) = -M1<x - 0>0 + R1<x - a>1 - P<x - L >1
4. Determine the magnitude of the force, P
P  M  g
P  980.7  N
5. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = L,
where both are zero.
At x = L+, V = M = 0
V = R1  P = 0
R1  P
R1  981  N
M1  R1 L
M1  1275 m N
6. Define the range for x
M = M1  R1 L = 0
x  0  in 0.005  L  L
7. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
S ( x z)  if ( x  z 1 0 )
8. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the
effect of the singularity functions.
V ( x)  R1 S ( x 0  mm)  P S ( x L)
M ( x)  M1 S ( x 0  mm)  R1 S ( x 0  mm)  ( x  0  mm)  P S ( x L)  ( x  L)
9. Plot the shear and moment diagrams.
© 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,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-12-2
1000
Shear
Diagram
800
V ( x)
N
600
400
200
0
0
0.5
1
1.5
2
1.5
2
x
m
0
Moment
Diagram
 300
M ( x)
Nm
 600
 900
 1200
 1500
0
0.5
1
x
m
FIGURE 3-12B
Shear and Moment Diagrams for Problem 3-12
10. The maximum value of the shear force ocuurs throughout the distance from x = 0 to x = L and is
R1  981  N
11. Find the maximum value of the bending moment by determining the value of x where the shear is zero.
Inspection of the shear diagram shows that this occurs at x = 0.
Mmax  M ( 0  mm)
Mmax  1275 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,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-13-1
PROBLEM 3-13
Statement:
Given:
Determine the impact force and dynamic deflection that will result when a 100-kg person jumps
up 25 cm and lands back on the board. Assume the board weighs 19 kg and deflects 8.5 cm
statically when the person stands on it. Find the reaction forces and construct the shear and
moment diagrams for this dynamic loading. Determine the maximum shear force, maximum
moment, and their locations along the length of the board.
Total board length
b  2000 mm
Supported length
a  700  mm
Mass of board
mboard  19 kg
Static board deflection
δstat  85 mm
Mass of person
mperson  100  kg
Height of jump
h  250  mm
2000
1300 = L
Fi
M1
Assumptions: 1. The board can be modelled as a cantilever
beam with maximum shear and moment at the
edge of the support.
Solution:
1.
FIGURE 3-13A
Free Body Diagram for Problem 3-13
The person impacts the board upon landing. Thus, the board is the struck object and the person is the
striking object. To determine the force exerted by the person we will first need to know the impact correction
factor from equation (3.15).
1
1
η  0.94
mboard
(1)
3  mperson
We can now use equation (3.14) to determine the impact force, Fi,


Fi  mperson g   1 
3.
700
See Figure 3-13 and Mathcad file P0313.
η 
2.
R1
1
2  η h 

δstat 
Fi  3.487  kN
(2)
Write an equation for the load function in terms of equations 3.17 and integrate the resulting function twice
using equations 3.18 to obtain the shear and moment functions. Note use of the unit doublet function to
represent the moment at the wall. For the beam in Figure 3-13,
q(x) = -M1<x - 0>-2 + R1<x - 0>-1 - Fi<x - l>-1
(3)
V(x) = -M1<x - 0>-1 + R1<x - 0>0 - Fi<x - l>0 + C1
(4)
M(x) = -M1<x - 0>0 + R1<x - 0>1 - Fi<x - l>1 + C1x+ C2
(5)
The reaction moment M1 at the wall is in the z direction and the forces R1 and Fi are in the y direction in
equation (4). All moments in equation (5) are in the z direction.
4.
Because the reactions have been included in the loading function, the shear and moment diagrams both
close to zero at each end of the beam, making C1 = C2 = 0.
5.
The reaction force R1 and the reaction moment M1 can be calculated from equations (4) and (5) respectively
by substituting the boundary conditions x = l+, V = 0, M = 0. Note that we can substitute l for l+ since their
difference is vanishingly small.
© 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,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-13-2
l  b  a
Unsupported beam length
l  1300 mm
V(l) = -M1<l - 0>-1 + R1<l - 0>0 - Fi<l - l>0 = 0
V = R1  Fi = 0
(6)
R1  Fi
R1  3.487  kN
M(l) = -M1<l - 0>0 + R1<l - 0>1 - Fi<l - l>1 = 0
M = M1  R1 l  Fi ( l  l) = 0
(7)
M1  R1 l
M1  4533 N  m
6. To generate the shear and moment functions over the length of the beam, equations (4) and (5) must be
evaluated for a range of values of x from 0 to l, after substituting the above values of C1, C2, R1, and M1 in them.
For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than the
dummy variable z, and a value of one when it is greater than or equal to z. It will have the same effect as the
singularity function.
Range of x
x  0  in 0.005  l  l
Unit step function
S ( x z)  if ( x  z 1 0 )
Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get
the effect of the singularity functions.
0
V ( x)  R1 S ( x 0  in)  ( x  0 )  Fi S ( x l)  ( x  l)
0
0
1
M ( x)  M1 S ( x 0  in)  ( x  0 )  R1 S ( x 0  in)  ( x  0 )  Fi S ( x l )  ( x  l)
(8)
1
Plot the shear and moment diagrams (see below).
Shear Diagram
Moment Diagram
0
1
3
V ( x)
kN
M ( x)
2
kN  m
3
1
0
2
4
0
0.5
1
5
0
0.5
1
x
x
m
m
1.5
7. The graphs show that the shear force and the moment are both largest at x = 0. The function values of
these points can be calculated from equations (4) and (5) respectively by substituting x = 0 and evaluating the
singularity functions:
Vmax = V(0) = R1<0 - 0>0 - Fi<0 - l>0 = R1
(9)
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
Vmax  R1
3-13-3
Vmax  3.49 kN
M.max = M(0) = -M1<0 - 0>0 + R1<0 - 0>1 - Fi<0 - l>1 = -M1
Mmax  M1
(10)
Mmax  4533 N  m
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-14-1
PROBLEM 3-14
Statement:
Units:
Given:
Figure P3-5 shows a child's toy called a pogo stick. The child stands on the pads, applying half
her weight on each side. She jumps off the ground, holding the pads up against her feet, and
bounces along with the spring cushioning the impact and storing energy to help each rebound.
Find the natural frequency of the system, the static deflection of the spring with the child
standing still, and the dynamic force and deflection when the child lands after jumping 2 in off
the ground.
blob :=
lbf ⋅ sec
2
in
Child's weight
Wc := 60⋅ lbf
Spring constant
k := 100⋅ lbf ⋅ in
Pogo stick weight
Wp := 5 ⋅ lbf
Height of drop
h := 2 ⋅ in
−1
Assumptions: 1. An approximate energy method will be acceptable.
2. The correction factor for energy dissipation will be applied.
Solution:
See Figure 3-14 and Mathcad file P0314.
Fi /2
1. Find the natural frequency of the (child/spring) system.
Mass of child (striker)
m :=
Mass of stick (struck)
mb :=
Natural frequency
ω :=
f :=
Wc
Fi /2
m = 0.155⋅ blob
g
Wp
mb = 0.013⋅ blob
g
k
ω = 25.367⋅
m
rad
sec
P
ω
2⋅ π
f = 4.037⋅ Hz
FIGURE 3-14
Free Body Diagram for Problem 3-14
2. The static deflection of the spring with the child standing still is
Static deflection of spring
δst :=
Wc
δst = 0.6⋅ in
k
3. Determine the mass ratio correction factor from equation (3.15):
Correction factor
1
η :=
1+
mb
η = 0.973
3⋅ m
4. Using equation (3.14), determine the dynamic force.


Fi := Wc⋅  1 +
1+
2 ⋅ η⋅ h 

δst 
Fi = 224⋅ lbf
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-15-1
PROBLEM 3-15
Statement:
A pen plotter imparts a constant acceleration of 2.5 m/sec2 to the pen assembly, which travels in
a straight line across the paper. The moving pen assembly weighs 0.5 kg. The plotter weighs 5
kg. What coefficient of friction is needed between the plotter feet and the table top on which it
sits to prevent the plotter from moving when the pen accelerates?
Given:
Acceleration of pen ass'y
a  2.5 m sec
Mass of pen ass'y
mpen  0.5 kg
mplot  5  kg
Mass of plotter
Solution:
1.
2
See Mathcad file P0315.
The force imparted to the pen assembly by the internal drive mechanism must be reacted at the table top by
the plotter feet. The horizontal force at the feet will be equal to the force on the pen assembly and must be
less than or equal to the maximum friction force, which is the product of the coefficient of friction and the
normal force, which is the weight of the plotter.
Horizontal driving
force on pen ass'y
Fpen  mpen a
Fpen  1.25 N
Weight of plotter
Wplot  mplot  g
Wplot  49.033 N
Minimum coefficient
of friction
μ 
Fpen
Wplot
μ  0.025
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-16-1
PROBLEM 3-16
Statement:
A track to guide bowling balls is designed with two round rods as shown in Figure P3-6. The
rods are not parallel to one another but have a small angle between them. The balls roll on the
rods until they fall between them and drop onto another track. The angle between the rods is
varied to cause the ball to drop at different locations. Each rod's unsupported length is 30 in
and the angle between them is 3.2 deg. The balls are 4.5 in dia and weigh 2.5 lb. The center
distance between the 1-in-dia rods is 4.2 in at the narrow end. Find the distance from the
narrow end at which the ball drops through and determine the worst-case shear and moment
maximum for the rods as the ball rolls a distance from the narrow end that is 98% of the distance
to drop. Assume that the rods are simply supported at each end and have zero deflection under
the applied loading. (Note that assuming zero deflection is unrealistic. This assumption will be
relaxed in the next chapter after deflection has been discussed.)
Given:
Unsupported rod length
Half-angle between rods
Bowling ball diameter
Solution:
See Figure 3-16 and Mathcad file P0316.
1.
L  30 in
α  1.6 deg
D  4.5 in
Bowling ball weight
Rod diameter
Half width of rod gap
Calculate the distance between the ball and rod
centers.
Distance between centers
h 
D d
2
A
W  2.5 lbf
d  1.0 in
c  2.1 in
A
h  2.75 in
c
TOP VIEW

F
u

x
W/2
F
SECTION A-A
width(x)
(a) Distance between the roll axis and
the rod axis.
(b) Partial FBD of the bowling ball.
FIGURE 3-16
Dimensions and Free Body Diagrams for Problem 3-16
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
2.
3-16-2
Let x be the distance along the roll axis, and u be the corresponding distance to the point of contact between
the ball and rods, measured along the rods. Then the distance from the center plane of the ball to the center
of a rod as shown in Figure 3-16(a) is,
width( x)  c cos( α)  x sin( α)
(1)
And the distance from the narrow end to the point at which the ball drops (assuming rigid rods) is
xdrop 
h  c cos( α)
xdrop  23.31  in
sin( α)
The distance along the rod corresponding to xdrop is
u drop 
3.
xdrop  h  sin( α)
The angle made by a line through the ball-rod centers and the horizontal plane (see Figure 3-16b) is
θ ( x)  acos

4.
width( x) 
h


When x = 0, this is
θ0  θ ( 0  in)
θ0  40.241 deg
When x = 0.98xdrop, this is
θ98%  θ  0.98 xdrop
θ98%  5.577  deg
The loading on the ball is symmetric about its center plane along the x-axis. Figure 3-16(b) shows a FBD of
one half of the ball with the internal forces along the plane of symmetry due to the reaction at the other rod
omitted. With these forces omitted we may only sum forces in the vertical direction.
 Fy :
F  sin( θ )  μ  F  cos( θ ) 
F=
5.
u drop  23.24  in
cos( α)
W
2
(2)
=0
W
(3)
2  ( sin( θ )  μ  cos( θ ) )
The ball will drop through the rods when  is zero. If there were no friction force present ( = 0) then F would
become very large as  approached zero. The presence of the friction term in the denominator of equation (3)
limits F to finite values. However, with the assumption that the rods are rigid, there is no way for the rods to
provide a normal force when  reaches zero. Thus, we will need to limit the range of  for this analysis.
Let
μ  0
Then
xmax 
u max 
Fmax 
and
θmin  θ98%
h  cos θmin  c cos( α)
sin( α)
xmax  h  sin( α)
cos( α)
W
2  sin θmin
xmax  22.84  in
u max  22.77  in
Fmax  12.86  lbf
(4)
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6.
3-16-3
Determine the worst-case shear and moment maximum for the rods as the ball rolls along their length from
Figure B-2(a) in Appendix B where a in the figure is u max. Then,


Mmax  Fmax u max  1 
u max 
L


Mmax  70.6 in lbf
(5)
For the shear, we must find the reactions, which are


R1  Fmax  1 
R2  Fmax  R1
u max 
L


R1  3.10 lbf
R2  9.76 lbf
The maximum absolute value of shear is the larger of these two. Thus
Vmax  R2
Vmax  9.8 lbf
(6)
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-17-1
PROBLEM 3-17
Statement:
A pair of ice tongs is shown in Figure P3-7. The ice weighs 50 lb and is 10 in wide across the
tongs. The distance between the handles is 4 in, and the mean radius r of a tong is 6 in. Draw
free-body diagrams of the two tongs and find all forces acting on them. Determine the bending
moment at point A.
Given:
Weight of ice
Distances
W  50 lbf
a x  11.0 in
a y  6.0 in
b x  5.0 in
b y  12.0 in
cx  2.0 in
cy  3.5 in
Assumptions: Assume that the horizontal force at C (the handle) is zero, thus Fc  0  lbf
Solution:
(1)
See Figure 3-17 and Mathcad file P0317.
F
F
F
C
FC
O
11.0 = ax
3.5 = cy
FO
2.0 = cx
A
12.0 = by
5.0 = bx
FB
B
W/2
W
FIGURE 3-17A
Free Body Diagrams for Problem 3-17
1.
Summing forces and moments on a single tong (see FBD above right).
 Fx
FO  FB  FC = 0
 Fy
W
 MC
FO cy  FB  b y  cy 
2
(2)
F=0
(3)
W
2
  b x  cx = 0
2.
From equations (1) and (2), FO = FB
3.
Eliminating FO from equations (4) and (5) and solving for FB
FB 
W   b x  cx
2 by
(4)
(5)
FB  14.58  lbf
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-17-2
F
4. From equation (3), the vertical force
on one handle is
F 
W
FC
F  25 lbf
2
O
5. From Figure 3-17B we see that, at any
section that we might take through the
tong, there will be an internal moment,
shear force, and axial force present. The
bending moment will be a maximum at
point A because it is the fartherest point
from the centroid of the system.
Summing forces and moments:
 Fx
C
-FDs cos  + FDn sin 
(6)
+ FO = 0
11.0 = ax
A
3.5 = cy
FO
2.0 = cx

D
FDs M D FDn
FIGURE 3-17B
Free Body Diagram with section at D for Problem 3-17
 Fy
-FDs sin  - FDn cos 
+F=0
(7)
 MO
F cx - M D - (FDs cos + FDn sin )(ay + rc sin )
+ (FDs sin + FDn cos )[ax - rc (1 - cos)] = 0
(8)
6.
Solving equations (6) and (7) for FDs and FDn
FDn = F  cos( θ )  FO sin( θ )
7.
FDs =
FDn sin( θ )  FO
cos( θ )
The maximum value of MD will occur at  = 0 deg. At  = 0 deg,
FO  FB
FDn  F
FDn  25 lbf
FDs  FO
FDs  14.58  lbf
MD  F  cx  FDs a y  FDn a x
MD  237.5  lbf  in
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-18-1
PROBLEM 3-18
Statement:
A tractor-trailer tipped over while negotiating an on-ramp to the New York Thruway. The road
has a 50-ft radius at that point and tilts 3 deg toward the outside of the curve. The 45-ft-long by
8-ft-wide by 8.5-ft-high trailer box (13 ft from ground to top) was loaded 44 415 lb of paper rolls in
two rows by two high as shown in Figure P3-8. The rolls are 40-in-dia by 38-in-long and weigh
about 900 lb each. They are wedged against rolling backward but not against sliding sidewards.
The empty trailer weighed 14 000 lb. The driver claims that he was traveling at less than 15 mph
and that the load of paper shifted inside the trailer, struck the trailer sidewall, and tipped the
truck. The paper company that loaded the truck claims the load was properly stowed and would
not shift at that speed. Independent test of the coefficient of friction between similar paper rolls
and a similar trailer floor give a value of 0.43 +/- 0.08. The composite center of gravity of the
loaded trailer is estimated to be 7.5 ft above the road. Determine the truck speed that would
cause the truck to just begin to tip and the speed at which the rolls will just begin to slide
sidways. What do you think caused the accident?
Given:
Weight of paper
Wp := 44415⋅ lbf
Weight of trailer
Wt := 14000⋅ lbf
Radius of curve
Nominal coefficient of friction
r := 50⋅ ft
μnom := 0.43
Coefficient of friction uncertainty
u μ := 0.08
Trailer width
Height of CG from pavement
w := 8 ⋅ ft
h := 7.5⋅ ft
Assumptions: 1. The paper rolls act as a monolith since they are tightly strapped together with steel bands.
2. The tractor has about 15 deg of potential roll freedom versus the trailer due to their relative angle
in plan during the turn combined with the substantial pitch freedom in the fifth wheel. So the trailer
can tip independently of the tractor.
3. The outside track width of the trailer tires is equal to the width of the trailer.
Solution:
See Figure 3-18 and Mathcad file P0318.
1. First, calculate the location of the
trailer's CG with respect to the outside
wheel when it is on the reverse-banked
curve. From Figure 3-18A,
Tilt angle
θ := 3 ⋅ deg
a := h ⋅ tan( θ )
a = 0.393⋅ ft
b :=
w
2
−a
xbar := b ⋅ cos( θ )
ybar := b⋅ sin( θ ) +
3°
b = 3.607⋅ ft
7.500'
h
cos( θ )
ybar = 7.699⋅ ft
a
The coordinates of the CG of the
loaded trailer with respect to the lower
outside corner of the tires are:
xbar = 3.602⋅ ft
ybar
xbar = 3.602⋅ ft
ybar = 7.699⋅ ft
b
xbar
4.000'
FIGURE 3-18A
Location of CG for Problem 3-18
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-18-2
2. The trailer is on the verge of tipping over when the copule due to centrifugal force is equal to the couple formed
by the weight of the loaded trailer acting through its CG and the vertical reaction at the outside edge of the tires.
At this instant, it is assumed that the entire weight of the trailer is reacted at the outside tires with the inside tires
carrying none of the weight. Any increase in tangential velocity of the tractor and trailer will result in tipping.
Summing moments about the tire edge (see Figure 3-18B),
ΣM
Fw⋅ xbar − Fc⋅ ybar = 0
(1)
where Fc is the centrifugal force due
to the normal acceleration as the tractor
and trailer go through the curve. The
normal acceleration is
a tip =
vtip
Fc
2
(2)
r
and the force necessary to keep the
tractor trailer following a circular path
is
Fc = mtot⋅ a tip
Fw
ybar
(3)
Rx
where mtot is the total mass of the
trailer and its payload. Combining
equations (2) and (3) and solving for
vtip, we have
vtip =
Fc⋅ r
Ry
xbar
(4)
mtot
FIGURE 3-18B
FBD of Trailer on the Verge of Tipping
or,
vtip =
3.
Fc⋅ r⋅ g
(5)
Fw
Calculate the minimum tipping velocity of the tractor/trailer. From equations (1) and (5),
Total weight
Fw := Wt + Wp
Centrifugal force required
to tip the trailer
Fc :=
Minimum tipping speed
vtip :=
xbar
ybar
⋅ Fw
Fc⋅ r⋅ g
Fw
Fw = 58415⋅ lbf
Fc = 27329⋅ lbf
vtip = 18.7⋅ mph
Thus, with the assumptions that we have made, the trailer would not begin to tip over until it reached a speed of
vtip = 18.7⋅ mph
4.
The load will slip when the friction force between the paper rolls and the trailer floor is no longer sufficient to
react the centrifugal force on the paper rolls. Looking at the FBD of the paper rolls in Figure 3-18C, we see that
Normal force between
paper and floor
Fn = Wp⋅ cos( θ ) − Fcp ⋅ sin( θ )
(6)
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-18-3
Tangential force tending to slide the paper
Ft = Wp⋅ sin( θ ) + Fcp ⋅ cos( θ )
(7)
Fcp
Centrifugal force on the paper
Fcp =
Wp
g
Wp
2
⋅ as =
W p vs
⋅
g r
(8)
Ft
Fn
But, the maximum friction force is
(9)
Ff = μ ⋅ Fn = Ft
FIGURE 3-18C
FBD of Paper on the Verge of Sliding
Substituting equation (9) into (7), then combining (6) and (7) to eliminate Fn, and solving for Fcp yields
Fcp =
Wp⋅ ( μ ⋅ cos( θ ) − sin( θ ) )
(10)
μ ⋅ sin( θ ) + cos( θ )
Substituting equation (10) into (8), to eliminate Fcp , and solving for vs yields
vs =
5.
 ( μ ⋅ cos( θ ) − sin( θ ) ) 
 μ ⋅ sin( θ ) + cos( θ )  ⋅ r⋅ g


(11)
Use the upper and lower limit on the coefficient of friction to determine an upper and lower limit on the speed
necessary to cause sliding.
Maximium coefficient
μmax := μnom + u μ
μmax = 0.51
Minimium coefficient
μmin := μnom − uμ
μmin = 0.35
Maximum velocity to cause sliding
vsmax :=
 ( μmax ⋅ cos( θ ) − sin( θ ) )
 μ ⋅ sin( θ ) + cos( θ )  ⋅ r⋅ g
 max

vsmax = 18.3⋅ mph
Minimum velocity to cause sliding
vsmin :=
6.
 ( μmin ⋅ cos( θ ) − sin( θ ) )
 μ ⋅ sin( θ ) + cos( θ )  ⋅ r⋅ g
 min

vsmin = 14.8⋅ mph
This very rough analysis shows that , if the coefficient of friction was at or near the low end of its measured
value, the paper load could slide at a tractor/trailer speed of 15 mph, which would lead to the trailer tipping over.
In any case, it appears that the paper load would slide before the truck would tip with the load in place.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-19-1
PROBLEM 3-19
Statement:
Assume that the CG of the paper rolls in Problem 3-18 is 2.5 ft above the floor of the trailer. At what
speed on the same curve will the pile of rolls tip over (not slide) with respect to the trailer?
Given:
Weight of paper
Wp := 44415⋅ lbf
Radius of curve
Paper roll length
Height of CG from floor
r := 50⋅ ft
L := 38⋅ in
h := 2.5⋅ ft
L = 3.167⋅ ft
Assumptions: The paper rolls act as a single, lumped mass and tip about one corner where they are braced
against sliding. The brace provides no moment support.
Solution:
See Figure 3-19 and Mathcad file P0319.
1. First, calculate the location of the paper's CG with
respect to the outside corner when it is on the
reverse-banked curve. From Figure 3-19,
Tilt angle
θ := 3⋅ deg
a := h⋅ tan ( θ)
a = 0.131⋅ ft
b := L − a
b = 3.036⋅ ft
xbar := b ⋅ cos ( θ)
xbar = 3.031⋅ ft
ybar := b⋅ sin ( θ) +
Fcp
2.500'
Wp
a
ybar
Rx
b
xbar R y
3.167'
h
FIGURE 3-19
cos ( θ)
FBD of Paper on the Verge of Tipping
ybar = 2.662⋅ ft
The coordinates of the CG of the paper with respect to the lower outside corner are:
xbar = 3.031⋅ ft
2.
ybar = 2.662⋅ ft
The paper is on the verge of tipping over when the couple due to centrifugal force is equal to the couple formed
by the weight of the paper acting through its CG and the vertical reaction at the outside edge of the rolls. At this
instant, it is assumed that the entire weight of the paper is reacted at the outside corner. Any increase in
tangential velocity of the tractor and trailer will result in tipping. Summing moments about the outside corner
nearest the floor (see Figure 3-19),
ΣM
Wp⋅ xbar − Fcp ⋅ ybar = 0
(1)
where Fcp is the centrifugal force due to the normal acceleration as the tractor and trailer go through the curve.
The normal acceleration is
a tip =
vtip
2
(2)
r
and the force necessary to keep the tractor trailer following a circular path is
Fcp = mp⋅ a tip
(3)
where mp is the mass of the paper. Combining equations (2) and (3) and solving for vtip, we have
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
vtip =
3-19-2
Fcp ⋅ r
(4)
mp
or,
vtip =
3.
Fcp ⋅ r⋅ g
(5)
Wp
Calculate the minimum paper tipping velocity of the tractor/trailer. From equations (1) and (5),
Centrifugal force required
to tip the paper
Fcp :=
Minimum tipping speed
vtip :=
xbar
ybar
⋅ Wp
Fcp ⋅ r⋅ g
Wp
Fcp = 50574⋅ lbf
vtip = 29.2⋅ mph
Thus, with the assumptions that we have made, the paper would not begin to tip over until the tractor/trailor
reached a speed of
vtip = 29.2⋅ mph
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-20-1
PROBLEM 3-20
Statement:
Assume that the load of paper rolls in Problem 3-18 will slide sideways at a truck speed of 20 mph
on the curve in question. Estimate the impact force of the cargo against the trailer wall. The
force-deflection characteristic of the trailer wall has been measured as approximately 400 lb/in.
Given:
Weight of paper
Wp := 44415⋅ lbf
Weight of trailer
Wt := 14000⋅ lbf
Speed of tractor/trailer
vt := 20⋅ mph
Radius of curve
Trailer width
Paper roll length
r := 50⋅ ft
w := 8 ⋅ ft
L := 38⋅ in
lbf
k := 400
in
Trailer wall stiffness
L = 3.167⋅ ft
Assumptions: 1. The paper rolls act as a monolith since they are tightly strapped together with steel bands.
2. The worst case will result if friction between the floor and the paper is neglected.
Solution:
1.
See Figure P3-8 and Mathcad file P0320.
Calculate the distance that the rolls will slide before impacting the wall.
s :=
2.
1
2
vt
2
r
2
vi = 64.266⋅
in
sec
With the paper as the moving mass and the trailer as the stationary or struck mass, calculate the correction factor
using equation (3.15)
1
1+
η = 0.905
Wt
3 ⋅ Wp
Calculate the static deflection caused by the paper against the trailer wall.
δst :=
6.
sec
2⋅ a p⋅ s
η :=
5.
in
a p = 206.507⋅
From elementary particle dynamics, estimate the velocity at impact due to the centripetal acceleration
vi :=
4.
s = 10⋅ in
Determine the centripetal acceleration at 20 mph.
a p :=
3.
⋅ ( w − 2⋅ L)
Wp
δst = 111.037⋅ in
k
Using equation (3.12), estimate the dynamic force of the paper rolls impacting the trailer wall.
Fi := Wp⋅ vi⋅
η
g ⋅ δst
Fi = 13114⋅ lbf
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-21-1
PROBLEM 3-21
Statement:
Figure P3-9 shows an automobile wheel with two common styles of lug wrench being used to
tighten the wheel nuts, a single-ended wrench in (a), and a double-ended wrench in (b). In each
case two hands are required to provide forces respectively at A and B as shown. The distance
between points A and B is 1 ft in both cases. The wheel nuts require a torque of 70 ft-lb. Draw
free body diagrams for both wrenches and determine the magnitudes of all forces and moments
on each wrench. Is there any difference between the way these two wrenches perform their
assigned task? Is one design better than the other? If so, why? Explain.
Given:
Distance between A and B
Tightening torque
d AB := 1 ⋅ ft
T := 70⋅ ft ⋅ lbf
Assumptions: 1. The forces exerted by the user's hands lie in a plane through the wrench that is also parallel to
the plane of the wheel.
2. The applied torque is perpendicular to the plane of the forces.
3. By virtue of 1 and 2 above, this is a planar problem that can be described in a 2D FBD.
Solution:
See Figure 3-21 and Mathcad file P0321.
1. Summing moments about the left end of the
wrench (for either case)
12" = dAB
F
T − F ⋅ d AB = 0
2. Solving for F
T
F :=
T
d AB
F = 70⋅ lbf
F
(a) Single-ended Wrench
3. This result is the same for both wrenches.
12" = dAB
Is there any difference between the way these
two wrenches perform their assigned task?
No, they both require the same
two-handed exertion of 70 lb from each
hand.
Is one design better than the other? If so,
why? Explain.
Design (b) has advantages over (a)
because it is balanced about the wheel
nut. This allows the user to spin the
wrench once the nut is loosened. It is
also slightly easier to apply the upward
and downward forces (F) in a plane with
design (b).
F
6"
T
F
(b) Double-ended Wrench
FIGURE 3-21
Free Body Diagrams for Problem 3-21
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-22-1
PROBLEM 3-22
Statement:
A roller-blade skate is shown in Figure P3-10. The polyurethane wheels are 72 mm dia. The
skate-boot-foot combination weighs 2 kg. The effective "spring rate" of the person-skate
subsystem is 6000 N/m. Find the forces on the wheels' axles for a 100-kg person landing a 0.5-m
jump on one foot. (a) Assume all 4 wheels land simultaneously. (b) Assume that one wheel
absorbs all the landing force.
Given:
Mass of struck member
Msys  2  kg
Stiffness of struck member
k  6000
N
Mass of striking member
m
Mperson  100  kg
Height of drop
h  0.5 m
Assumptions: Equation (3.14) applies in this case.
Solution:
1.
See Figure P3-10 and Mathcad file P0322.
The weight of the striking mass is
Wperson  Mperson g
2.
The static deflection of the subsystem is
δst 
3.
Wperson
δst  163.444  mm
k
The correction factor is
1
η 
1
4.
Wperson  980.7  N
η  0.993
Msys
3  Mperson
From equation (3.14), the force of impact is

Fi   1 

1
2  η h 
  Wperson
δst 
Fi  3.59 kN
(a) If this will be absorbed by 4 wheel axles, the force per axle is
Fa 
Fi
Fa  897  N
4
(b) If one wheel absorbs all force
Fb  Fi
Fb  3.59 kN
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-23a-1
PROBLEM 3-23a
Statement:
Given:
A beam is supported and loaded as shown in Figure P3-11a. Find the reactions, maximum shear,
and maximum moment for the data given in row a from Table P3-1.
Beam length
L  1  m
Distance to distributed load
a  0.4 m
L
b
a
Distance to concentrated load b  0.6 m
1
Distributed load magnitude
w  200  N  m
Concentrated load
F  500  N
F
w
R2
R1
Solution:
See Figures 3-23 and Mathcad file P0323a.
FIGURE 3-23A
Free Body Diagram for Problem 3-23
1. From inspection of Figure P3-11a, write the load function equation
q(x) = R1<x - 0>-1 - w<x - 0>0 + w<x - a>0 - F<x - b>-1 + R2<x - L>-1
2. Integrate this equation from - to x to obtain shear, V(x)
V(x) = R1<x - 0>0 - w<x - 0>1 + w<x - a>1 - F<x - b>0 + R2<x - L>0
3. Integrate this equation from - to x to obtain moment, M(x)
M(x) = R1<x - 0>1 - w<x - 0>2/2 + w<x - a>2/2 - F<x - b>1 + R2<x - L>1
4. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = L,
where both are zero.
At x = L+, V = M = 0
V = R1  w ( L)  w ( L  a )  F  R2 = 0
M = R 1 L 
R1 
w
2
L 
w
2
2
L 
F
L
w
2
2
 ( L  a)  F  ( L  b) = 0
 ( L  b) 
w
2 L
 ( L  a)
2
R1  264  N
R2  w a  F  R1
5. Define the range for x
R2  316  N
x  0  m 0.005  L  L
6. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than
z, and a value of one when it is greater than or equal to z.
S ( x z)  if ( x  z 1 0 )
7. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get
the effect of the singularity functions.
V ( x)  R1 S ( x 0  m)  w S ( x 0  m)  ( x)  w S ( x a )  ( x  a )  F  S ( x b )  R2 S ( x L)
M ( x)  R1 S ( x 0  m)  x 
w
2
2
 S ( x 0  m)  x 
w
2
2
 S ( x a )  ( x  a )  F  S ( x b )  ( x  b )
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-23a-2
8. Plot the shear and moment diagrams.
400
Shear
Diagram
200
V ( x)
0
N
 200
 400
0
0.2
0.4
0.6
0.8
x
m
Moment
Diagram
150
100
M ( x)
Nm
50
0
0
0.2
0.4
0.6
0.8
x
m
FIGURE 3-23aB
Shear and Moment Diagrams for Problem 3-23a
9. Determine the maximum shear and maximum moment from inspection of the diagrams.
Maximum shear:
Vmax  V ( b )
Vmax  316  N
Maximum moment occurs where V is zero, which is x = b:
Mmax  M ( b )
Mmax  126.4  N  m
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-24a-1
PROBLEM 3-24a
Statement:
Given:
A beam is supported and loaded as shown in Figure P3-11b. Find the reactions, maximum shear,
and maximum moment for the data given in row a from Table P3-1.
Beam length
L  1  m
Distance to distributed load
a  0.4 m
a
1
w  200  N  m
Distributed load magnitude
F
w
F  500  N
Concentrated load
Solution:
L
See Figures 3-24 and Mathcad file P0324a.
1. From inspection of Figure P3-11b, write the load function
equation
M1
R1
FIGURE 3-24A
Free Body Diagram for Problem 3-24
q(x) = -M1<x - 0>-2 + R1<x - 0>-1 - w<x - a>0 - F<x - L>-1
2. Integrate this equation from - to x to obtain shear, V(x)
V(x) = -M1<x - 0>-1 + R1<x - 0>0 - w<x - a>1 - F<x - L>0
3. Integrate this equation from - to x to obtain moment, M(x)
M(x) = -M1<x - 0>0 + R1<x - 0>1 - w<x - a>2/2 - F<x - L>1
4. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = L,
where both are zero.
At x = L+, V = M = 0
V = R1  [ w  ( L  a )  F ] = 0
M = M1  R1 L 
w
2
2
 ( L  a) = 0
R1  w ( L  a )  F
M1 
w
2
5. Define the range for x
R1  620  N
2
 ( L  a )  R1 L
M1  584  N  m
x  0  m 0.005  L  L
6. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than
z, and a value of one when it is greater than or equal to z.
S ( x z)  if ( x  z 1 0 )
7. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get
the effect of the singularity functions.
V ( x)  R1 S ( x 0  m)  w S ( x a )  ( x  a )  F  S ( x L)
M ( x)  M1  R1 S ( x 0  m)  x 
w
2
2
 S ( x a )  ( x  a )  F  S ( x L)  ( x  L)
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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,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-24a-2
8. Plot the shear and moment diagrams.
Shear
Diagram
600
V ( x)
N
400
200
0
0
0.2
0.4
0.6
0.8
0.6
0.8
x
m
Moment
Diagram
0
 150
M ( x)
Nm
 300
 450
 600
0
0.2
0.4
x
m
FIGURE 3-24aB
Shear and Moment Diagrams for Problem 3-24a
9. Determine the maximum shear and maximum moment from inspection of the diagrams.
Maximum shear:
Vmax  V ( 0  m)
Vmax  620  N
Maximum moment occurs where V is zero, which is x = 0:
Mmax  M ( 0  m)
Mmax  584  N  m
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-25a-1
PROBLEM 3-25a
Statement:
Given:
A beam is supported and loaded as shown in Figure P3-11d. Find the reactions, maximum shear,
and maximum moment for the data given in row a from Table P3-1.
Beam length
L  1  m
Distance to distributed load
a  0.4 m
L
b
a
Distance to concentrated load b  0.6 m
F
1
Distributed load magnitude
w  200  N  m
Concentrated load
F  500  N
w
R2
R1
Solution:
See Figures 3-25 and Mathcad file P0325a.
FIGURE 3-25A
Free Body Diagram for Problem 3-25
1. From inspection of Figure P3-11c, write the load function equation
q(x) = R1<x - 0>-1 - w<x - a>0 + R2<x - b>-1 - F<x - L>-1
2. Integrate this equation from - to x to obtain shear, V(x)
V(x) = R1<x - 0>0 - w<x - a>1 + R2<x - b>0 - F<x - L>0
3. Integrate this equation from - to x to obtain moment, M(x)
M(x) = R1<x - 0>1 - w<x - a>2/2 + R2<x - b>1 - F<x - L>1
4. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = L,
where both are zero.
At x = L+, V = M = 0
V = R1  w  ( L  a )  R2  F = 0
M = R 1 L 
R1 
w
2
2
 ( L  a )  R2 ( L  b ) = 0
w
2
   ( L  a )  F  ( L  b )  w ( L  a )  ( L  b )
b 2

1
R2  w ( L  a )  F  R1
5. Define the range for x
R1  353  N
R2  973  N
x  0  m 0.005  L  L
6. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than
z, and a value of one when it is greater than or equal to z.
S ( x z)  if ( x  z 1 0 )
7. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get
the effect of the singularity functions.
V ( x)  R1 S ( x 0  m)  w S ( x a )  ( x  a )  R2 S ( x b )  F  S ( x L)
M ( x)  R1 S ( x 0  m)  x 
w
2
2
 S ( x a )  ( x  a )  R2 S ( x b )  ( x  b )
© 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.
MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-25a-2
8. Plot the shear and moment diagrams.
1
Shear
Diagram
0.5
V ( x)
0
kN
 0.5
1
0
0.2
0.4
0.6
0.8
x
m
Moment
Diagram
0
 75
M ( x)
Nm
 150
 225
 300
0
0.2
0.4
0.6
0.8
x
m
FIGURE 3-25aB
Shear and Moment Diagrams for Problem 3-25a
9. Determine the maximum shear and maximum moment from inspection of the diagrams.
Maximum shear:
Vmax  V ( b )
Vmax  580.0  N
Maximum moment occurs where V is zero, which is x = a:
Mmax  M ( b )
Mmax  216  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.
MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-26a-1
PROBLEM 3-26a
Statement:
Given:
Solution:
A beam is supported and loaded as shown in Figure P3-11d. Find the reactions, maximum shear,
and maximum moment for the data given in row a from Table P3-1.
Beam length
L  1  m
Distance to distributed load
a  0.4 m
Distance to reaction load
b  0.6 m
L
b
a
1
Distributed load magnitude
w  200  N  m
Concentrated load
F  500  N
F
w
R2
R1
See Figures 3-26 and Mathcad file P0326a.
FIGURE 3-26A
Free Body Diagram for Problem 3-26
1. From inspection of Figure 3-26aA, write the load function equation
q(x) = R1<x - 0>-1 - w<x - a>0 + R2<x - b>-1 - F<x - a>-1
2. Integrate this equation from - to x to obtain shear, V(x)
V(x) = R1<x - 0>0 - w<x - a>1 + R2<x - b>0 - F<x - a>0
3. Integrate this equation from - to x to obtain moment, M(x)
M(x) = R1<x - 0>1 - w<x - a>2/2 + R2<x - b>1 - F<x - a>1
4. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = L,
where both are zero.
At x = L+, V = M = 0
V = R1  w  ( L  a )  R2  F = 0
M = R 1 L 
R1 
w
2
2
 ( L  a )  R2 ( L  b )  F ( L  a ) = 0
w
2
   ( L  a )  F  ( b  a )  w ( L  a )  ( L  b )
b 2

1
R2  w ( L  a )  F  R1
5. Define the range for x
R1  147  N
R2  473  N
x  0  m 0.005  L  L
6. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than
z, and a value of one when it is greater than or equal to z.
S ( x z)  if ( x  z 1 0 )
7. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get
the effect of the singularity functions.
V ( x)  R1 S ( x 0  m)  w S ( x a )  ( x  a )  R2 S ( x b )  F  S ( x a )
M ( x)  R1 S ( x 0  m)  x 
w
2
2
 S ( x a )  ( x  a )  R2 S ( x b )  ( x  b )  F  S ( x a )  ( x  a )
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-26a-2
8. Plot the shear and moment diagrams.
500
Shear
Diagram
250
V ( x)
0
N
 250
 500
0
0.2
0.4
0.6
0.8
x
m
Moment
Diagram
60
40
M ( x)
Nm
20
0
 20
0
0.2
0.4
0.6
0.8
x
m
FIGURE 3-26aB
Shear and Moment Diagrams for Problem 3-26a
9. Determine the maximum shear and maximum moment from inspection of the diagrams.
Maximum shear:
Vmax  V ( b  0.001  mm)
Vmax  393  N
Maximum moment occurs where V is zero, which is x = a:
Mmax  M ( a )
Mmax  58.7 N  m
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-27-1
PROBLEM 3-27
Statement:
A storage rack is to be designed to hold the paper roll of Problem 3-8 as shown in Figure
P3-12. Determine the reactions and draw the shear and moment diagrams for the mandrel that
extends 50% into the roll.
Given:
Paper roll dimensions
OD  1.50 m
ID  0.22 m
Lroll  3.23 m
Roll density
ρ  984  kg m
3
Assumptions: 1. The paper roll's weight creates a concentrated load acting at the tip of the mandrel.
2. The mandrel's root in the stanchion experiences a distributed load over its length of engagemen
Solution:
See Figure 3-27 and Mathcad file P0327.
W
1. Determine the weight of the roll
and the length of the mandrel.
W 

4
π
2
2

 OD  ID  Lroll  ρ  g
W  53.9 kN
M1
Lm
R1
Lm  0.5 Lroll
FIGURE 3-27
Lm  1.615  m
Free Body Diagram for Problem 3-27
2. From inspection of Figure 3-27, write the load function equation
q(x) = -M1<x - 0>-2 + R1<x - 0>-1 - W<x - L>-1
3. Integrate this equation from - to x to obtain shear, V(x)
V(x) = -M1<x - 0>-1 + R1<x - 0>0 - W<x - L>0
4. Integrate this equation from - to x to obtain moment, M(x)
M(x) = -M1<x - 0>0 + R1<x - 0>1 - W<x - L>1
5. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = L,
where both are zero.
At x = L+, V = M = 0
V = R1  W = 0
M = M1  R1 L = 0
R1  W
R1  53.895 kN
M1  R1 Lm
M1  87.040 kN  m
6. Define the range for x
x  0  m 0.005  Lm  Lm
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-27-2
7. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than
z, and a value of one when it is greater than or equal to z.
S ( x z)  if ( x  z 1 0 )
8. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get
the effect of the singularity functions.
V ( x)  R1 S ( x 0  m)  W  S  x Lm
M ( x)  M1  R1 S ( x 0  m)  x  W  S  x Lm   x  Lm
9. Plot the shear and moment diagrams.
Shear
Diagram
40
V ( x)
kN
20
0
0
0.5
1
1.5
2
x
m
Moment
Diagram
20
1.615
 10
M ( x)
kN  m
 40
 70
 100
0
0.5
1
1.5
2
x
m
FIGURE 3-27B
Shear and Moment Diagrams for Problem 3-27
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-28-1
PROBLEM 3-28
Statement:
Figure P3-13 shows a forklift truck negotiating a 15 deg ramp to to drive onto a 4-ft-high loading
platform. The truck weighs 5 000 lb and has a 42-in wheelbase. Determine the reactions and draw
the shear and moment diagrams for the worst case of loading as the truck travels up the ramp.
Given:
Ramp angle
Platform height
Truck weight
Truck wheelbase
θ  15 deg
h  4  ft
W  5000 lbf
Lt  42 in
h  48 in
Assumptions: 1. The worst case is when the truck CG is located at the center of the beam's span.
2. Use a coordinate frame that has the x-axis along the long axis of the beam.
3. Ignore traction forces and the weight components along the x-axis of the beam.
4. There are two ramps, one for each side of the forklift.
5. The location of the CG in Figure P3-13 is 32 in from the front wheel and 10 in from the rear
wheel.
CGa  32 in
Solution:
CGb  10 in
See Figure 3-28 and Mathcad file P0328.
L
b
a
CG a
y
CG b
R1

Fa
Wa
Fb
x
Wb
R2
FIGURE 3-28A
Dimensions and Free Body Diagram for Problem 3-28
1. Determine the length of the beam between supports and the distances a and b.
Length of beam
With the CG at midspan, we have
and
L 
h
L  15.455 ft
sin( θ )
a  CGa =
L
2
a 
L
b 
L
2
2
 CGa
a  5.061  ft
 CGb
b  8.561  ft
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-28-2
2. The weight distribution on the wheels is determined from the distance from the front wheel to the CG. Each
wheel weight is divided by 2 to get the weight on a single ramp.
Weight on front wheel
Wa 
CGb W

Lt 2
Wa  595  lbf
Weight on rear wheel
Wb 
W
Wb  1905 lbf
2
 Wa
3. The normal force on the ramp at each wheel is adjusted for the ramp angle.
Load at front wheel
Fa  Wa cos( θ )
Fa  575  lbf
Load at rear wheel
Fb  Wb cos( θ )
Fb  1840 lbf
4. From inspection of Figure 3-28A, write the load function equation
q(x) = R1<x - 0>-1 - Fa<x - a>-1 - Fb<x - b>-1 + R2<x - L>-1
5. Integrate this equation from - to x to obtain shear, V(x)
V(x) = R1<x - 0>0 - Fa<x - a>0 - Fb<x - b>0 + R2<x - L>0
6. Integrate this equation from - to x to obtain moment, M(x)
M(x) = R1<x - 0>1 - Fa<x - a>1 - Fb<x - b>1 + R2<x - L>1
7. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = L,
where both are zero.
At x = L+, V = M = 0
V = R1  Fa  Fb  R2 = 0
M = R1 L  Fa ( L  a )  Fb ( L  b ) = 0
R1 
1
L
 Fa ( L  a )  Fb ( L  b )
R2  Fa  Fb  R1
8. Define the range for x
R1  1207 lbf
R2  1207 lbf
x  0  m 0.005  L  L
9. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than
z, and a value of one when it is greater than or equal to z.
S ( x z)  if ( x  z 1 0 )
10. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the
effect of the singularity functions.
V ( x)  R1 S ( x 0  m)  Fa S ( x a )  Fb S ( x b )  R2 S ( x L)
M ( x)  R1 S ( x 0  m)  x  Fa S ( x a )  ( x  a )  Fb S ( x b )  ( x  b )  R2 S ( x L)  ( x  L)
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-28-3
11. Plot the shear and moment diagrams.
2000
Shear
Diagram
1000
V ( x)
0
lbf
 1000
 2000
0
2
4
6
8
10
12
14
16
x
ft
Moment
Diagram
10000
15.455
8000
M ( x)
6000
ft  lbf
4000
2000
0
0
2
4
6
8
10
12
14
16
x
ft
FIGURE 3-28B
Shear and Moment Diagrams for Problem 3-28
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-29-1
PROBLEM 3-29
_____
Statement:
Run the TKSolver or Mathcad model for Case Study 1A and move the point of application of the
hand force along the lever by changing the values of Rb2, recalculate and observe the changes to
the forces and moments.
Problem:
Determine the forces on the elements of the bicycle brake lever assembly shown in Figure 3-1
during braking.
Given:
The geometry of each element is known. The average human's hand can develop a grip force of
about 267 N (60 lb) in the lever position shown.
Magnitude of handle force Fb2
Fb2 := 267⋅ N
Direction of handle force Fb2
θb2 := 270⋅ deg
Direction of cable force Fc2
θc2 := 184⋅ deg
Direction of cable force Fcable
θcable := 180⋅ deg
Position vector components (Change the value of Rb2x and note the results)
Rb2x := 19⋅ mm
Rc2x := −25⋅ mm
R12x := −12⋅ mm
Rb2y := −4⋅ mm
Rc2y := 0⋅ mm
R12y := −7⋅ mm
R21x := 7⋅ mm
Rb1x := 47.5⋅ mm
R31x := −27⋅ mm
R21y := 19⋅ mm
Rb1y := −14⋅ mm
R31y := 30⋅ mm
Assumptions: The accelerations are negligible. All forces are coplanar and two-dimensional. A class 1 load
model is appropriate and a static analysis is acceptable. The higher applied load will be used as a
worst case, assuming that it can be reached before bottoming the tip of the handle on the handgrip.
If that occurs, it will change the beam's boundary conditions and the analysis.
Solution:
1.
See Figures 3-1, 3-2, and Mathcad file P0329.
Figure 3-1 shows the hand brake lever assembly, which consists of three subassemblies: the handlebar (1), the
lever (2), and the cable (3). The lever is pivoted to the handlebar and the cable is connected to the lever. The
cable runs within a plastic-lined sheath (for low friction) down to the brake caliper assembly at the bicycle's
wheel rim. The user's hand applies equal and opposite forces at some point on the lever and handgrip. These
forces are transformed to a larger force in the cable by reason of the lever ratio of part 2.
Figure 3-1 is a free-body diagram of the entire assembly since it shows all the forces and moments acting on it
except for its weight, which is small compared to the applied forces and is thus neglected for this analysis. The
"broken away" portion of the handlebar provides internal x and y force components and a moment. These are
arbitrarily shown as positive in sign. Their actual signs will "come out in the wash" in the calculations. The
known applied forces are shown in their actual directions and senses.
2.
Figure 3-2 shows the three subassembly elements separated and drawn as free-body diagrams with all relevant
forces and moments applied to each element, again neglecting the weights of the parts. The lever (part 2) has
three forces on it, Fb2, Fc2, and F12. The two-character subscript notation used here should be read as, force of
element 1 on 2 (F12) or force at B on 2 (Fb2), etc. This defines the source of the forces (first subscript) and the
element on which it acts (second subscript).
This notation will be used consistently throughout this text for both forces and position vectors such as Rb2,
Rc2, and R12 in Figure 3-2, which serve to locate the above three forces in a local, non rotating coordinate system
whose origin is at the center of gravity (CG) of the element or subassembly being analyzed. (See foot note on
page 83 of the text).
On this brake lever, Fb2 is an applied force whose magnitude and direction are known. Fc2 is the force in the
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-29-2
cable. Its direction is known but not its magnitude. Force F12 is provided by part 1 on part 2 at the pivot pin.
Its magnitude and direction are both unknown. We can write equations 3.3b for this element to sum forces
in the x and y directions and sum moments about the CG. Note that all unknown reactive forces and moments are
initially assumed positive in the equations. Their true signs will come out in the calculation. (See foot note on
page 84 of the text).
ΣFx = Fb2x + Fc2x + F12x = 0
(a)
ΣFy = Fb2y + Fc2y + F12y = 0
ΣMz = ( R12 × F12) + ( Rb2 × Fb2) + ( Rc2 × Fc2 ) = 0
The cross products in the moment equation represent the "turning forces" or moments created by the
application of these forces at points remote from the CG of the element. Recall that these cross products can be
expanded to
ΣMz = ( R12x⋅ F12y − R12y⋅ F12x) ...  = 0
+ ( Rb2x⋅ Fb2y − Rb2y⋅ Fb2x) ...
+ ( R ⋅ F − R ⋅ F ) 
 c2x c2y c2y c2x 
(b)
We have three equations and four unknowns (F12x, F12y, Fc2x, Fc2y) at this point, so we need another equation. It
is available from the fact that the direction of Fc2 is known. (The cable can pull only along its axis). We can
express one component of the cable force Fc2 in terms of its other component and the known angle θc2 of the
cable.
(c)
Fc2y = Fc2x⋅ tan( θc2 )
We will now use a Mathcad solve block to solve equations a through c.
Calculate components of Fb2
Fb2x := Fb2⋅ cos( θb2)
Fb2x = −0 ⋅ N
Fb2y := Fb2⋅ sin( θb2)
Fb2y = −267⋅ N
Guess
F12x := 1000⋅ N
Given
Fb2x + Fc2x + F12x = 0
Fc2x := −1000⋅ N
F12y := 1000⋅ N
Fc2y := −1000⋅ N
Fb2y + Fc2y + F12y = 0
( R12x⋅ F12y − R12y⋅ F12x) ...  = 0
+ ( Rb2x⋅ Fb2y − Rb2y⋅ Fb2x) ...
+ ( R ⋅ F − R ⋅ F ) 
 c2x c2y c2y c2x 
Fc2y = Fc2x⋅ tan( θc2 )
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-29-3
 F12x 
F 
 12y  := Find ( F , F , F , F )
12x 12y c2x c2y
 Fc2x 


 Fc2y 
Components of the unknown forces F12, and Fc2
F12x = 1047⋅ N
3.
Fc2x = −1047⋅ N
F12y = 340⋅ N
Fc2y = −73.2⋅ N
Part 3 in Figure 3-2 is the cable that passes through a hole in part 1. This hole is lined with a low friction material,
which allows us to assume no friction at the joint between parts 1 and 3. We will further assume that the three
forces F13, Fc3, and Fcable form a concurrent system of forces acting through the CG and thus create no moment.
With this assumption, only a summation of forces is necessary for this element.
ΣFx = Fcablex + F13x + Fc3x = 0
(d)
ΣFy = Fcabley + F13y + Fc3y = 0
Using Newton's third law, we have Fc3x := −Fc2x and Fc3y := −Fc2y.
We also assume that the cable entering from the left is horizontal and that the reaction F13 is vertical, thus
Fcabley := 0 ⋅ N
and
(e)
F13x := 0 ⋅ N
We can now solve for the forces on part 3 directly,
Fcablex := −F13x − Fc3x
Fcablex = −1047⋅ N
F13y := −Fcabley − Fc3y
F13y = −73.2⋅ N
The assembly of elements labeled part 1 in Figure 3-2 has both force and moments on it (i.e., it is not a concurrent
system), so the three equations 3.3b are needed.
ΣFx = F21x + Fb1x + F31x + Px + Fsheathx = 0
(f)
ΣFy = F21y + Fb1y + F31y + Py = 0
ΣMz = Mh + ( R21 × F21) + ( Rb1 × Fb1) + ( R31 × F31) ... = 0
+ ( Rp × Fp) + ( Rd × Fsheath )
Expanding cross products in the moment equation gives the moment magnitude as
ΣMz = Mh + ( R21x⋅ F21y − R21y⋅ F21x) ... = 0
+ ( Rb1x⋅ Fb1y − Rb1y⋅ Fb1x) ...
+ ( R31x⋅ F31y − R31y⋅ F31x) ...
+ ( R ⋅ F − R ⋅ F ) ...
 Px Py Py Px
+ ( 0 − Rdy ⋅ Fsheathx)





(g)
Using Newton's third law, we have
F31x := −F13x
F21x := −F12x
Fb1x := −Fb2x
(h)
F31y := −F13y
F21y := −F12y
Fb1y := −Fb2y
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-29-4
Fsheathx := −Fcablex
Given
RPx := −27⋅ mm
RPy := 0 ⋅ mm
Rdx := −41⋅ mm
Rdy := 27⋅ mm
We will now use a Mathcad solve block to solve equations (f) through (h).
Guess
Px := 1000⋅ N
Given
F21x + Fb1x + F31x + Px + Fsheathx = 0
Mh := −100⋅ N ⋅ m
Py := 0 ⋅ N
F21y + Fb1y + F31y + Py = 0
Mh + ( R21x⋅ F21y − R21y⋅ F21x) ... = 0
+ ( Rb1x⋅ Fb1y − Rb1y⋅ Fb1x) ... 
+ ( R31x⋅ F31y − R31y⋅ F31x) ... 
+ ( R ⋅ P − R ⋅ P ) ...

 Px y Py x

+ ( 0⋅ N ⋅ m − Rdy⋅ Fsheathx)

 Px 
 
 Py  := Find( Px , Py , Mh)
M 
 h
Summarizing, the results obtained for a grip force Fb2 = 267⋅ N are:
Handlebar (1)
Fb1x = 0 ⋅ N
Fb1y = 267⋅ N
F21x = −1047⋅ N
F21y = −340⋅ N
F31x = 0 ⋅ N
F31y = 73.2⋅ N
−6
Px = 1 × 10
⋅N
Py = 0 ⋅ N
Mh = 0.0⋅ N ⋅ m
Lever (2)
Cable (3)
Fc2x = −1047⋅ N
Fc2y = −73.2⋅ N
F12x = 1047⋅ N
F12y = 340⋅ N
Fc3x = 1047⋅ N
Fc3y = 73.2⋅ N
F13x = 0 ⋅ N
F13y = −73.2⋅ N
Fcablex = −1047⋅ N
Fcabley = 0 ⋅ N
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-30-1
PROBLEM 3-30
_____
Statement:
Run the TKSolver or Mathcad model for Case Study 2A and move the point of application of the
crimp force along the jaw by changing the values of Rhand, recalculate and observe the changes to
the forces and moments.
Problem:
Determine the forces on the elements of the crimping tool shown in Figure 3-3 during a crimp
operation.
Given:
The geometry is known and the tool develops a crimp force of 2000 lb (8896 N) at closure in the
position shown.
Applied crimp force
Fc4x := −1956.30⋅ lbf
Fc4y := 415.82⋅ lbf
Position vector components (Change the value of Rhand and note the results)
Rc4x := 0.454⋅ in
R12x := 1.399⋅ in
R32x := 2.199⋅ in
Rc4y := 0.337⋅ in
R12y := 0.049⋅ in
R32y := 0.077⋅ in
R23x := −0.602⋅ in
R43x := 0.602⋅ in
R14x := −0.161⋅ in
R23y := 0.127⋅ in
R43y := −0.127⋅ in
R14y := −0.758⋅ in
R34x := 0.161⋅ in
R34y := 0.758⋅ in
Rhand := −4.40⋅ in
Assumptions: The accelerations are negligible. All forces are coplanar and two-dimensional. A class 1 load
model is appropriate and a static analysis is acceptable.
Solution:
1.
See Figures 3-3, 3-4, and Mathcad file P0330.
Figure 3-3 shows the tool in the closed position, in the process of crimping a metal connector onto a wire. The
user's hand provides the input forces between links 1 and 2, shown as the reaction pair Fhand. The user can grip
the handle anywhere along its length but we are assuming a nominal moment arm of Rhand for the application of
the resultant of the user's grip force (see Figure 3-4). The high mechanical advantage of the tool transforms the
grip force to a large force at the crimp.
Figure 3-3 is a free-body diagram of the entire assembly, neglecting the weight of the tool, which is small
compared to the crimp force. There are four elements, or links, in the assembly, all pinned together. Link 1 can
be considered to be the "ground" link, with the other links moving with respect to it as the jaw is closed. The
desired magnitude of the crimp force Fc is defined and its direction will be normal to the surfaces at the crimp.
2.
Figure 3-4 shows the elements of the crimping tool assembly separated and drawn as free-body diagrams with all
forces applied to each element, again neglecting their weights as being insignificant compared to the applied
forces. The centers of gravity of the respective elements are used as the origins of the local, non rotating
coordinate systems in which the points of application of all forces on the element are located. (See footnote on
page 116 of the text).
3.
We will consider link 1 to be the ground plane and analyze the remaining moving links. Note that all unknown
forces and moments are initially assumed positive. Link 4 has three forces acting on it: Fc4 is the known (desired)
force at the crimp, and F14 and F34 are the reaction forces from links 1 and 3, respectively. The magnitudes of
these two forces are unknown as is the direction of F14. The direction of F34 will be the same as link 3, since it is
a two-force member. Writing equations 3.3b for this element:
ΣFx = F14x + F34x + Fc4x = 0
ΣFy = F14y + F34y + Fc4y = 0
(a)
ΣMz = ( R14x⋅ F14y − R14y⋅ F14x) ...  = 0
+ ( R34x⋅ F34y − R34y⋅ F34x) ...
+ ( R ⋅ F − R ⋅ F ) 
 c4x c4y c4y c4x 
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-30-2
We have three equations and four unknowns (F14x, F14y, F34x, F34y) at this point, so we need another equation.
It is available from the fact that the direction of F34 is known. We can express one component of the force F34 in
terms of its other component and the known angle θ3 of link 3.
(b)
F34y = F34x⋅ tan( θ3)
(c)
where
θ3 := 168⋅ deg
Guess
F14x := 500⋅ lbf
Given
F14x + F34x + Fc4x = 0
F34x := 1000⋅ lbf
F14y := −100⋅ lbf
F34y := −100⋅ lbf
F14y + F34y + Fc4y = 0
( Rc4x⋅ Fc4y − Rc4y⋅ Fc4x) ...  = 0
+ ( R14x⋅ F14y − R14y⋅ F14x) ...
+ ( R ⋅ F − R ⋅ F ) 
 34x 34y 34y 34x 
F34y = F34x⋅ tan( θ3)
 F14x 
F 
 14y  := Find ( F , F , F , F )
14x 14y 34x 34y
 F34x 


 F34y 
Components of the unknown forces F14, and F34
F14x = 442.9⋅ lbf
4.
5.
F14y = −94.1⋅ lbf
F34x = 1513.4⋅ lbf
F34y = −321.7⋅ lbf
Link 3 has two forces on it, F23 and F43. Because this is a two-force link, these two forces are equal in magnitude
and opposite in direction. Also, from Newton's third law, F43 = - F34. Thus,
F43x := −F34x
F43y := −F34y
F23x := −F43x
F23y := −F43y
F43x = −1513.4⋅ lbf
F43y = 321.7⋅ lbf
F23x = 1513.4⋅ lbf
F23y = −321.7⋅ lbf
(d)
Link 2 has three forces acting on it: Fhand is the unknown force from the hand, and F12 and F32 are the reaction
forces from links 1 and 3, respectively. Force F12 is provided by part 1 on part 2 at the pivot pin and force F32 is
provided by part 3 acting on part 2 at their pivot pin. The magnitude and direction of F32 is known and the
direction of Fhand is known. Using equations 3.3b, we can solve for the magnitude of Fhand and the two
components of F12. From the third law,
F32x := −F23x
F32y := −F23y
F32x = −1513.4⋅ lbf
F32y = 321.7⋅ lbf
ΣFx = F12x + F32x = 0
(e)
ΣFy = Fhand + F12y + F32y = 0
ΣMz = ( R12 × F12) + ( R32 × F32) ... = 0
+ ( Rhand × Fhand )
Guess
F12x := 1500⋅ lbf
F12y := −100⋅ lbf
Fhand := 100⋅ lbf
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
Given
3-30-3
F12x + F32x = 0
F12y + F32y + Fhand = 0
( R12x⋅ F12y − R12y⋅ F12x) ...  = 0
+ ( R32x⋅ F32y − R32y⋅ F32x) ...
+ R ⋅ F

 hand hand

 F12x 


 F12y  := Find( F12x , F12y , Fhand )
F 
 hand 
F12x = 1513.4⋅ lbf
6.
7.
F12y = −373.4⋅ lbf
Fhand = 51.7⋅ lbf
The four forces on link 1 can now be determined using the third law.
F21x := −F12x
F21y := −F12y
F41x := −F14x
F41y := −F14y
F21x = −1513.4⋅ lbf
F21y = 373.4⋅ lbf
F41x = −442.9⋅ lbf
F41y = 94.1⋅ lbf
Fc1x := −Fc4x
Fc1y := −Fc4y
Fc1x = 1956.3⋅ lbf
Fc1y = −415.8⋅ lbf
The solution to this problem for the scaled dimensions in Figure 3-3 assuming a 2000-lb (8896-N) force applied at
the crimp, normal to the crimp surface, is given above. The total forces at the pivot points are:
F12 :=  F12x + F12y
2
0.5
Pivot A
F32 :=  F32x + F32y
2
0.5
Pivot B
F43 :=  F43x + F43y
2
0.5
Pivot C
F14 :=  F14x + F14y
2
0.5
Pivot D
2
2
2
2

F12 = 1559⋅ lbf

F32 = 1547⋅ lbf

F43 = 1547⋅ lbf

F14 = 453⋅ lbf
The moment that must be applied to the handles to generate the crimp force of
Crimp force
Fc4 :=  Fc4x + Fc4y
Moment
Mh := Rhand ⋅ Fhand
2
2
0.5

Fc4 = 2000⋅ lbf
Mh = 227⋅ lbf ⋅ in
This moment can be obtained with a force of Fhand = 52⋅ lbf applied at mid-handle. This force is within the
physiological grip-force capacity of the average human.
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-31-1
PROBLEM 3-31
_____
Statement:
Run the TKSolver or Mathcad model for Case Study 2A and move the point of application of P
along the x direction by changing the values of Rpx, recalculate and observe the changes to the
forces and moments. What happens when the vertical force P is centered on link 3? Also, change
the angle of the applied force P to create an x component and observe the effects on the forces and
moments on the elements.
Problem:
Determine the forces on the elements of the scissors-jack in the position shown in Figure 3-5.
Given:
The geometry is known and the jack supports a force of 1000 lb (4448 N) in the position shown.
Support force
Px := 0.0⋅ lbf
Py := −1000⋅ lbf
Position vector components (Change the value of Rpx and note the results)
Rpx := −0.50⋅ in
R12x := −3.12⋅ in
R32x := 2.08⋅ in
Rpy := 0.87⋅ in
R12y := −1.80⋅ in
R32y := 1.20⋅ in
R42x := 2.71⋅ in
R23x := −0.78⋅ in
R43x := 0.78⋅ in
R42y := 1.00⋅ in
R23y := −0.78⋅ in
R43y := −0.78⋅ in
R14x := 3.12⋅ in
R24x := −2.58⋅ in
R34x := −2.08⋅ in
R14y := −1.80⋅ in
R24y := 1.04⋅ in
R34y := 1.20⋅ in
Angle of gear teeth common normal
θ := −45.0⋅ deg
Assumptions: The accelerations are negligible. The jack is on level ground. The angle of the elevated car chassis
does not impart an overturning moment to the jack. All forces are coplanar and two-dimensional.
A class 1 load model is appropriate and a static analysis is acceptable.
Solution:
See Figures 3-5 through 3-8, and Mathcad file P0331.
1.
Figure 3-5 shows a schematic of a simple scissors jack used to raise a car. It consists of six links that are
pivoted and/or geared together and a seventh link in the form of a lead screw that is turned to raise the jack.
While this is clearly a three-dimensional device, it can be analyzed as a two-dimensional one if we assume that
the applied load (from the car) and the jack are exactly vertical (in the z direction). If so, all forces will be in the xy
plane. This assumption is valid if the car is jacked from a level surface. If not, then there will be some forces in
the yz and xz planes as well. The jack designer needs to consider the more general case, but for our simple
example we will initially assume two-dimensional loading. For the overall assembly as shown in Figure 3-5, we
can solve for the reaction force Fg, given force P, by summing forces: Fg = -P.
2.
Figure 3-6 shows a set of free-body diagrams for the entire jack. Each element or subassembly of interest has
been separated from the others and the forces and moments shown acting on it (except for its weight, which is
small compared to the applied forces and is thus neglected for this analysis). The forces and moments can be
either internal reactions at interconnections with other elements or external loads from the "outside world." The
centers of gravity of the respective elements are used as the origins of the local, non rotating coordinate systems
in which the points of application of all forces on the element are located. In this design, stability is achieved by
the mating of two pairs of crude (non involute) gear segments acting between links 2 and 4 and between links 5
and 7. These interactions are modeled as forces acting along a common normal shared by the two teeth. This
common normal is perpendicular to the common tangent at the contact point.
There are 3 second-law equations available for each of the seven elements allowing 21 unknowns. An additional
10 third-law equations will be needed for a total of 31. This is a cumbersome system to solve for such a simple
device, but we can use its symmetry to advantage in order to simplify the problem.
3.
Figure 3-7 shows the upper half of the jack assembly. Because of the mirror symmetry between the upper
and lower portions, the lower half can be removed to simplify the analysis. The forces calculated for this half will
be duplicated in the other. If we wished, we could solve for the reaction forces at A and B using equations 3.3b
from this free-body diagram of the half-jack assembly.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-31-2
4.
Figure 3-8a shows the free-body diagrams for the upper half of the jack assembly, which are essentially the same
as those of Figure 3-6. We now have four elements but can consider the subassembly labeled 1 to be the
"ground," leaving three elements on which to apply equations 3.3. Note that all forces and moments are initially
assumed positive in the equations.
5.
Link 2 has three forces acting on it: F42 is the unknown force at the gear tooth contact with link 4; F12 and F32 are
the unknown reaction forces from links 1 and 3, respectively. Force F12 is provided by part 1 on part 2 at the
pivot pin and force F32 is provided by part 3 acting on part 2 at their pivot pin. The magnitudes and the
directions of these pin forces and the magnitude of F42 are unknown. The direction of F42 is along the common
normal shown in Figure 3-8b. Write equations 3.3b for this element to sum the forces in the x and y directions
and sum moments about the CG (with the cross products expanded).
ΣFx = F12x + F32x + F42x = 0
(a)
ΣFy = F12y + F32y + F42y = 0
ΣMz = ( R12x⋅ F12y − R12y⋅ F12x) ...  = 0
+ ( R32x⋅ F32y − R32y⋅ F32x) ...
+ ( R ⋅ F − R ⋅ F ) 
 42x 42y 42y 42x 
6.
Link 3 has three forces acting on it: P, F23 and F43. Only P is known. Writing equations 3.3b for this element
gives
ΣFx = F23x + F43x + Px = 0
(b)
ΣFy = F23y + F43y + Py = 0
ΣMz = ( R23x⋅ F23y − R23y⋅ F23x) ...  = 0
+ ( R43x⋅ F43y − R43y⋅ F43x) ...
+ ( R ⋅ P − R ⋅ P )

 px y py x

7.
Link 4 has three forces acting on it: F24 is the unknown force from link 2; F14 and F34 are the unknown reaction
forces from links 1 and 3, respectively.
ΣFx = F14x + F24x + F34x = 0
(c)
ΣFy = F14y + F24y + F34y = 0
ΣMz = ( R14x⋅ F14y − R14y⋅ F14x) ...  = 0
+ ( R24x⋅ F24y − R24y⋅ F24x) ...
+ ( R ⋅ F − R ⋅ F ) 
 34x 34y 34y 34x 
8.
9.
The nine equations in sets a through c have 16 unknowns in them, F12x, F12y, F32x, F32y, F23x, F23y, F43x, F43y,
F14x , F14y, F34x, F34y, F24x, F24y, F42x, F42y. We can write the third-law relationships between action-reaction pairs
at each of the joints to obtain six of the seven additional equations needed:
F32x = −F23x
F32y = −F23y
F34x = −F43x
F34y = −F43y
F42x = −F24x
F42y = −F24y
(d)
The last equation needed comes from the relationship between the x and y components of the force F24 (or
F42) at the tooth/tooth contact point. Such a contact (or half) joint can transmit force (excepting friction force)
only along the common normal , which is perpendicular to the joint's common tangent as shown in Figure 3-8b.
The common normal is also called the axis of transmission.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-31-3
The tangent of the angle of this common normal relates the two components of the force at the joint:
(e)
F24y = F24x⋅ tan( θ )
10. Equations (d) and (e) will be substituted into equations (a) through (c) to create a set of nine simultaneous
equations for solution.
Guess
Given
F12x := 500⋅ lbf
F12y := 500⋅ lbf
F14x := −500⋅ lbf
F23x := 500⋅ lbf
F23y := 500⋅ lbf
F24x := 500⋅ lbf
F43x := −500⋅ lbf
F43y := 500⋅ lbf
( R12x⋅ F12y − R12y⋅ F12x) ...
+ ( −R32x⋅ F23y + R32y⋅ F23x) ...
+ ( −R ⋅ F ⋅ tan( θ ) + R ⋅ F )
42x 24x
42y 24x

 =0



F14y := 500⋅ lbf
F12x − F23x − F24x = 0
F12y − F23y − F24x⋅ tan( θ ) = 0
( R23x⋅ F23y − R23y⋅ F23x) ...  = 0
+ ( R43x⋅ F43y − R43y⋅ F43x) ...
+ ( R ⋅ P − R ⋅ P )

 px y py x

F23x + F43x + Px = 0
( R14x⋅ F14y − R14y⋅ F14x) ...
 =0
+ ( R24x⋅ F24x⋅ tan( θ ) − R24y⋅ F24x) ...
+ ( −R ⋅ F + R ⋅ F )

34x 43y
34y 43x


F14x + F24x − F43x = 0
F23y + F43y + Py = 0
F14y + F24x⋅ tan( θ ) − F43y = 0
 F12x 


 F12y 
 F14x 


 F14y 
 F23x  := Find ( F , F , F , F , F , F , F , F , F )
12x 12y 14x 14y 23x 23y 24x 43x 43y


 F23y 
F 
 24x 
 F43x 


 F43y 
Results:
F14x = −877.8⋅ lbf
F14y = 469.6⋅ lbf
F24x = 290.1⋅ lbf
F24y := F24x⋅ tan( θ )
F34x := −F43x
F34y := −F43y
F23x = 587.7⋅ lbf
F23y = 820.5⋅ lbf
F43x = −587.7⋅ lbf
F43y = 179.5⋅ lbf
F12x = 877.8⋅ lbf
F12y = 530.4⋅ lbf
F32x := −F23x
F32y := −F23y
F42x := −F24x
F42y := −F24y
F24y = −290.1⋅ lbf
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-32-1
PROBLEM 3-32
_____
Statement:
Figure P3-14 shows a cam-follower arm. If the load P = 200 lb, what spring force is needed at the
right end to maintain a minimum load between cam and follower of 25 lb? Find the maximum shear
force and bending moment in the follower arm. Plot the shear and moment diagrams.
Given:
Load at left end of beam
Load at cam follower
P := 200⋅ lbf
Pcam := 25⋅ lbf
Distance from left end to: Pivot point
Cam follower
Spring
Solution:
1.
a := 10⋅ in
b := 22⋅ in
c := 29⋅ in
See Figure P3-14 and Mathcad file P0332.
Draw a FBD of the cam-follower arm (beam).
c
b
P
C
R
2.
Pcam
Fspring
From inspection of the FBD, write the load function equation
q(x) = -P<x - 0>-1 + R<x - a>-1 + Pcam<x - b >-1 - Fspring<x - 0>-1
3.
Integrate this equation from -∞ to x to obtain shear, V(x)
V(x) = -P<x - 0>0 + R<x - a>0 + Pcam<x - b >0 - Fspring<x - 0>0
4.
Integrate this equation from -∞ to x to obtain moment, M(x)
M(x) = -P<x - 0>1 + R<x - a>1 + Pcam<x - b >1 - Fspring<x - 0>1
5.
Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = c, where
both are zero.
At x = c+, V = M = 0
V = −P + R + Pcam − Fspring = 0
M = −P⋅ c + R⋅ ( c − a ) + Pcam⋅ ( c − b ) = 0
Fspring :=
P⋅ a + Pcam⋅ ( b − a )
c−a
R := Fspring + P − Pcam
6.
Define the range for x
x := 0 ⋅ in , 0.002⋅ c .. c
Fspring = 121.05⋅ lbf
R = 296.05⋅ lbf
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
7.
3-32-2
For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
S ( x , z) := if ( x ≥ z , 1 , 0 )
8.
Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the
effect of the singularity functions.
V ( x) := −P⋅ S ( x , 0 ⋅ in) + R⋅ S ( x , a ) + Pcam⋅ S ( x , b ) − Fspring⋅ S ( x , c)
M ( x) := −P⋅ S ( x , 0 ⋅ in) ⋅ x + R⋅ S ( x , a ) ⋅ ( x − a) + Pcam⋅ S ( x , b ) ⋅ ( x − b ) − Fspring⋅ S ( x , c) ⋅ ( x − c)
9.
Plot the shear and moment diagrams and find the maximum shear force and bending moment.
SHEAR DIAGRAM
200
100
V ( x)
lbf
0
− 100
− 200
− 300
0
10
20
30
x
in
Vmax := V ( 0 ⋅ in)
Vmax = 200⋅ lbf
MOMENT DIAGRAM
0
− 500
M ( x)
in⋅ lbf
− 1000
− 1500
− 2000
0
10
20
30
x
in
Mmax := M ( a )
Mmax = 2000⋅ in⋅ lbf
© 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.
MACHINE DESIGN - An Integrated Approach, 4th Ed.
PROBLEM 3-33
3-33-1
_____
Statement:
Write a computer program or equation solver model to calculate all the singularity functions listed
in equations 3.17. Set them up as functions that can be called from within any other program or
model.
Solution:
See Mathcad file P0333.
1.
No solution is provided for this programming problem.
© 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.
MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-34a-1
PROBLEM 3-34a
Statement:
Given:
A beam is supported and loaded as shown in Figure P3-15. Find the reactions, maximum shear, and
maximum moment for the data given in row a from Table P3-2.
Beam length
L := 20⋅ in
Distance to RH bearing
a := 16⋅ in
Distance to concentrated load
b := 18⋅ in
Concentrated load
P := 1000⋅ lbf
b
P
R2
R1
a
FIGURE 3-34aA
Solution:
1.
See Figure 3-34 and Mathcad file P0334a.
Free Body Diagram for Problem 3-34
From inspection of Figure 3-34, write the load function equation
q(x) = R1<x - 0>-1 + R2<x - b>-1 - P<x - L>-1
2.
Integrate this equation from -∞ to x to obtain shear, V(x)
V(x) = R1<x - 0>0 + R2<x - b>0 - P<x - L>0
3.
Integrate this equation from -∞ to x to obtain moment, M(x)
M(x) = R1<x - 0>1 + R2<x - b>1 - P<x - L>1
4.
Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = b, where
both are zero.
At x = b +, V = M = 0
V = R1 + R2 − P = 0
M = R1 ⋅ b + R2 ⋅ ( b − a ) = 0
R1 :=
P
a
⋅ (a − b)
R2 := P − R1
R1 = −125⋅ lbf
R2 = 1125⋅ lbf
5.
Define the range for x
6.
For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
x := 0 ⋅ m , 0.002⋅ L .. L
S ( x , z) := if ( x ≥ z , 1 , 0 )
7.
Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the
effect of the singularity functions.
V ( x) := R1⋅ S ( x , 0 ⋅ in) + R2⋅ S ( x , a ) − P⋅ S ( x , b )
M ( x) := R1⋅ S ( x , 0 ⋅ in) ⋅ x + R2⋅ S ( x , a ) ⋅ ( x − a ) − P⋅ S ( x , b ) ⋅ ( x − b)
8.
Plot the shear and moment diagrams.
© 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.
MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-34a-2
SHEAR DIAGRAM
MOMENT DIAGRAM
1000
1000
0
500
V ( x)
M ( x)
lbf
in⋅ lbf
− 1000
0
− 2000
− 500
0
5
10
15
− 3000
20
0
5
10
x
x
in
in
15
20
FIGURE 3-34aB
Shear and Moment Diagrams for Problem 3-34a
9.
Determine the maximum shear and maximum moment from inspection of the diagrams.
Maximum shear:
Vmax := V ( a )
Vmax = 1000⋅ lbf
Maximum moment occurs where V is zero, which is x = a:
Mmax := M ( a )
Mmax = 2000⋅ in⋅ lbf
© 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.
MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-35a-1
PROBLEM 3-35a
Statement:
Input data:
A beam is supported and loaded as shown in Figure P3-15. Write a computer program or equation
solver model to find the reactions and calculate and plot the loading, shear, and moment functions.
Test the program with the data given in row a from Table P3-2.
b
Enter data in highlighted areas
Beam length
L := 20⋅ in
Distance to RH bearing
a := 16⋅ in
Distance to concentrated load
b := 18⋅ in
Concentrated load
F := 1000⋅ lbf
F
R2
R1
a
FIGURE 3-34aA
Solution:
1.
See Figures 3-35 and Mathcad file P0335a.
Free Body Diagram for Problem 3-34
From inspection of Figure 3-35, write the load function equation
q(x) = R1<x - 0>-1 + R2<x - b>-1 - F<x - L>-1
2.
Integrate this equation from -∞ to x to obtain shear, V(x)
V(x) = R1<x - 0>0 + R2<x - b>0 - F<x - L>0
3.
Integrate this equation from -∞ to x to obtain moment, M(x)
M(x) = R1<x - 0>1 + R2<x - b>1 - F<x - L>1
4.
Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = b, where
both are zero.
At x = b +, V = M = 0
V = R1 + R2 − F = 0
M = R1 ⋅ b + R2 ⋅ ( b − a ) = 0
R1 :=
F
a
⋅ ( a − b)
R2 := F − R1
R1 = −125⋅ lbf
R2 = 1125⋅ lbf
5.
Define the range for x
6.
For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
x := 0 ⋅ m , 0.002⋅ L .. L
S ( x , z) := if ( x ≥ z , 1 , 0 )
7.
Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the
effect of the singularity functions.
V ( x) := R1⋅ S ( x , 0 ⋅ in) + R2⋅ S ( x , a ) − F ⋅ S ( x , b )
M ( x) := R1⋅ S ( x , 0 ⋅ in) ⋅ x + R2⋅ S ( x , a ) ⋅ ( x − a ) − F ⋅ S ( x , b ) ⋅ ( x − b )
8.
Plot the shear and moment diagrams.
© 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.
MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-35a-2
SHEAR DIAGRAM
MOMENT DIAGRAM
1000
1000
0
500
V ( x)
M ( x)
lbf
in⋅ lbf
− 1000
0
− 2000
− 500
0
5
10
15
− 3000
20
0
5
10
x
x
in
in
15
20
FIGURE 3-34aB
Shear and Moment Diagrams for Problem 3-35a
9.
Determine the maximum shear and maximum moment from inspection of the diagrams.
Maximum shear:
Vmax := V ( a )
Vmax = 1000⋅ lbf
Maximum moment occurs where V is zero, which is x = a:
Mmax := M ( a )
Mmax = 2000⋅ in⋅ lbf
© 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.
MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-36a-1
PROBLEM 3-36a
Statement:
Given:
Solution:
1.
A beam is supported and loaded as shown in Figure P3-16. Find the reactions, maximum shear, and
maximum moment for the data given in row a from Table P3-2.
Beam length
L := 20⋅ in
Distance to RH bearing
L := 20⋅ in
Distance to start of load
a := 16⋅ in
Distance to end of load
b := 18⋅ in
Distributed load
p := 1000⋅
b
a
L
R1
lbf
in
p
R2
FIGURE 3-36aA
See Figures 3-36 and Mathcad file P0336a.
Free Body Diagram for Problem 3-36
From inspection of Figure 3-36, write the load function equation
q(x) = R1<x - 0>-1 - p<x - a>0 + p<x - b>0 + R2<x - L>-1
2.
Integrate this equation from -∞ to x to obtain shear, V(x)
V(x) = R1<x - 0>0 - p<x - a>1 + p<x - b>1 + R2<x - L>0
3.
Integrate this equation from -∞ to x to obtain moment, M(x)
M(x) = R1<x - 0>1 - p<x - a>2/2 + p<x - b>2/2 + R2<x - L>1
4.
Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = L, where
both are zero.
At x = L+, V = M = 0
V = R1 − p ⋅ ( L − a ) + p ⋅ ( L − b ) + R2 = 0
M = R1 ⋅ L −
R1 :=
p
2⋅ L
p
2
2
⋅ (L − a) +
p
2
2
⋅ ( L − b ) + R2⋅ ( L − b) = 0
⋅ 2 ⋅ ( b − a ) ⋅ L + a − b
2
2

R1 = 300⋅ lbf
R2 := p ⋅ ( b − a ) − R1
R2 = 1700⋅ lbf
5.
Define the range for x
6.
For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
x := 0 ⋅ in , 0.002⋅ L .. L
S ( x , z) := if ( x ≥ z , 1 , 0 )
7.
Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the
effect of the singularity functions.
V ( x) := R1⋅ S ( x , 0 ⋅ in) − p ⋅ S ( x , a ) ⋅ ( x − a) + p ⋅ S ( x , b ) ⋅ ( x − b ) + R2⋅ S ( x , L)
M ( x) := R1⋅ S ( x , 0 ⋅ in) ⋅ x −
p
2
2
⋅ S(x , a)⋅ ( x − a) +
p
2
2
⋅ S ( x , b ) ⋅ ( x − b ) + R2⋅ S ( x , L) ⋅ ( x − L)
© 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.
MACHINE DESIGN - An Integrated Approach, 4th Ed.
8.
3-36a-2
Plot the shear and moment diagrams.
SHEAR DIAGRAM
MOMENT DIAGRAM
1000
5000
4000
0
V ( x)
M ( x)
lbf
in⋅ lbf
3000
2000
− 1000
1000
− 2000
0
5
10
15
20
0
0
5
10
x
x
in
in
15
20
FIGURE 3-36aB
Shear and Moment Diagrams for Problem 3-36a
9.
Determine the maximum shear and maximum moment from inspection of the diagrams.
Maximum shear:
Vmax := V ( b )
Vmax = 1700⋅ lbf
Maximum moment occurs where V is zero, which is x = c, where:
c :=
R1 ⋅ b + R2 ⋅ a
R1 + R2
Mmax := M ( c)
c = 16.3⋅ in
Mmax = 4845⋅ in⋅ lbf
© 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.
MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-37a-1
PROBLEM 3-37a
Statement:
Input data:
A beam is supported and loaded as shown in Figure P3-16. Write a computer program or equation
solver model to find the reactions and calculate and plot the loading, shear, and moment functions.
Test the program with the data given in row a from Table P3-2.
Enter data in highlighted areas
Beam length
Solution:
1.
b
L := 20⋅ in
Distance to RH bearing
L := 20⋅ in
Distance to start of load
a := 16⋅ in
Distance to end of load
b := 18⋅ in
Distributed load
p := 1000⋅
a
p
L
R1
R2
lbf
FIGURE 3-37aA
in
Free Body Diagram for Problem 3-37
See Figures 3-37 and Mathcad file P0337a.
From inspection of Figure 3-37, write the load function equation
q(x) = R1<x - 0>-1 - p<x - a>0 + p<x - b>0 + R2<x - L>-1
2.
Integrate this equation from -∞ to x to obtain shear, V(x)
V(x) = R1<x - 0>0 - p<x - a>1 + p<x - b>1 + R2<x - L>0
3.
Integrate this equation from -∞ to x to obtain moment, M(x)
M(x) = R1<x - 0>1 - p<x - a>2/2 + p<x - b>2/2 + R2<x - L>1
4.
Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = L, where
both are zero.
At x = L+, V = M = 0
V = R1 − p ⋅ ( L − a ) + p ⋅ ( L − b ) + R2 = 0
M = R1 ⋅ L −
R1 :=
p
2⋅ L
p
2
2
⋅ (L − a) +
p
2
2
⋅ ( L − b ) + R2⋅ ( L − b) = 0
⋅ 2 ⋅ ( b − a ) ⋅ L + a − b
2
2

R1 = 300⋅ lbf
R2 := p ⋅ ( b − a ) − R1
R2 = 1700⋅ lbf
5.
Define the range for x
6.
For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
x := 0 ⋅ in , 0.002⋅ L .. L
S ( x , z) := if ( x ≥ z , 1 , 0 )
7.
Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the
effect of the singularity functions.
V ( x) := R1⋅ S ( x , 0 ⋅ in) − p ⋅ S ( x , a ) ⋅ ( x − a) + p ⋅ S ( x , b ) ⋅ ( x − b ) + R2⋅ S ( x , L)
M ( x) := R1⋅ S ( x , 0 ⋅ in) ⋅ x −
p
2
2
⋅ S(x , a)⋅ ( x − a) +
p
2
2
⋅ S ( x , b ) ⋅ ( x − b ) + R2⋅ S ( x , L) ⋅ ( x − L)
© 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.
MACHINE DESIGN - An Integrated Approach, 4th Ed.
8.
3-37a-2
Plot the shear and moment diagrams.
SHEAR DIAGRAM
MOMENT DIAGRAM
1000
5000
4000
0
V ( x)
M ( x)
lbf
in⋅ lbf
3000
2000
− 1000
1000
− 2000
0
5
10
15
20
0
0
5
10
x
x
in
in
15
20
FIGURE 3-37aB
Shear and Moment Diagrams for Problem 3-37a
9.
Determine the maximum shear and maximum moment from inspection of the diagrams.
Maximum shear:
Vmax := V ( b )
Vmax = 1700⋅ lbf
Maximum moment occurs where V is zero, which is x = c, where:
c :=
R1 ⋅ b + R2 ⋅ a
R1 + R2
Mmax := M ( c)
c = 16.3⋅ in
Mmax = 4845⋅ in⋅ lbf
© 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.
MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-38a-1
PROBLEM 3-38a
Statement:
Given:
A beam is supported and loaded as shown in Figure P3-17. Find the reactions, maximum shear, and
maximum moment for the data given in row a from Table P3-2.
Beam length
b
L := 20⋅ in
Distance to RH bearing
a := 16⋅ in
Distance to concentrated load
b := 18⋅ in
Concentrated load
P := 1000⋅ lbf
Distributed load
p := 1000⋅ lbf ⋅ in
P
p
R2
R1
a
−1
FIGURE 3-38aA
Solution:
1.
2.
See Figure 3-38 and Mathcad file P0338a.
Free Body Diagram for Problem 3-38
Determine the distance from the origin to the left and right ends of the roller.
Distance to left end
e := 0.1⋅ a
e = 1.600⋅ in
Distance to right end
f := 0.9⋅ a
f = 14.400in
⋅
From inspection of Figure 3-38, write the load function equation
q(x) = R1<x - 0>-1 - p<x - e>0 + p<x - f>0 + R2<x - a>-1 - P<x - b>-1
3.
Integrate this equation from -∞ to x to obtain shear, V(x)
V(x) = R1<x - 0>0 - p<x - e>1 + p<x - f>1 + R2<x - a>0 - P<x - b>0
4.
Integrate this equation from -∞ to x to obtain moment, M(x)
M(x) = R1<x - 0>1 - p<x - e>2/2 + p<x - f>2/2 + R2<x - a>1 - P<x - b>1
5.
Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = b, where
both are zero.
At x = b +, V = M = 0
V = R1 − p ⋅ ( b − e ) + p ⋅ ( b − f ) + R2 − P = 0
M = R1 ⋅ b −
R1 :=
p
2
2
⋅ ( b − e) +
p
2
2
⋅ ( b − f ) + R2⋅ ( b − a) = 0
 e2 − f 2

b − a

+ f − e ⋅ p − 
⋅P
 2⋅ a

 a 
R2 := p ⋅ ( f − e) − R1 + P
R1 = 6275⋅ lbf
R2 = 7525⋅ lbf
6.
Define the range for x
7.
For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
x := 0 ⋅ m , 0.002⋅ L .. L
S ( x , z) := if ( x ≥ z , 1 , 0 )
8.
Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the
effect of the singularity functions.
V ( x) := R1⋅ S ( x , 0 ⋅ m) − p ⋅ S ( x , e) ⋅ ( x − e) + p ⋅ S ( x , f ) ⋅ ( x − f ) + R2⋅ S ( x , a ) − P⋅ S ( x , b )
p
2
2
⋅ S ( x , e) ⋅ ( x − e) + ⋅ S ( x , f ) ⋅ ( x − f ) ...
2
2
+ R2⋅ S ( x , a ) ⋅ ( x − a) − P⋅ S ( x , b ) ⋅ ( x − b )
M ( x) := R1⋅ S ( x , 0 ⋅ m) ⋅ x −
p
© 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.
MACHINE DESIGN - An Integrated Approach, 4th Ed.
9.
3-38a-2
Plot the shear and moment diagrams.
SHEAR DIAGRAM
MOMENT DIAGRAM
10000
30000
5000
20000
V ( x)
lbf
M ( x)
0
in⋅ lbf
10000
− 5000
− 10000
0
0
5
10
15
− 10000
20
0
5
10
x
x
in
in
15
20
FIGURE 3-38aB
Shear and Moment Diagrams for Problem 3-38a
9.
Determine the maximum shear and maximum moment from inspection of the diagrams.
Maximum shear:
Vmax := V ( f )
Vmax = 6525⋅ lbf
Maximum moment occurs where V is zero, which is x = c:
c−e
R1
=
f −c
R2 − P
Mmax := M ( c)
c :=
f ⋅ R1 + e ⋅ R2 − e ⋅ P
R1 + R2 − P
c = 7.875⋅ in
Mmax = 29728⋅ in⋅ lbf
© 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.
MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-39a-1
PROBLEM 3-39a
Statement:
Input data:
A beam is supported and loaded as shown in Figure P3-17. Write a computer program or equation
solver model to find the reactions and calculate and plot the loading, shear, and moment functions.
Test the program with the data given in row a from Table P3-2.
Enter data in highlighted areas
Beam length
L := 20⋅ in
Distance to RH bearing
a := 16⋅ in
Distance to concentrated load
b := 18⋅ in
Concentrated load
P := 1000⋅ lbf
Distributed load
Solution:
1.
2.
b
p := 1000⋅ lbf ⋅ in
P
p
R2
R1
a
−1
FIGURE 3-39aA
Free Body Diagram for Problem 3-39
See Figure 3-39 and Mathcad file P0339a.
Determine the distance from the origin to the left and right ends of the roller.
Distance to left end
e := 0.1⋅ a
e = 40.64⋅ mm
Distance to right end
f := 0.9⋅ a
f = 365.76mm
⋅
From inspection of Figure 3-39, write the load function equation
q(x) = R1<x - 0>-1 - p<x - e>0 + p<x - f>0 + R2<x - a>-1 - P<x - b>-1
3.
Integrate this equation from -∞ to x to obtain shear, V(x)
V(x) = R1<x - 0>0 - p<x - e>1 + p<x - f>1 + R2<x - a>0 - P<x - b>0
4.
Integrate this equation from -∞ to x to obtain moment, M(x)
M(x) = R1<x - 0>1 - p<x - e>2/2 + p<x - f>2/2 + R2<x - a>1 - P<x - b>1
5.
Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = b, where
both are zero.
At x = b +, V = M = 0
V = R1 − p ⋅ ( b − e ) + p ⋅ ( b − f ) + R2 − P = 0
M = R1 ⋅ b −
R1 :=
p
2
2
⋅ ( b − e) +
p
2
2
⋅ ( b − f ) + R2⋅ ( b − a) = 0

 e2 − f 2
b − a

+ f − e ⋅ p − 
⋅P
2
⋅
a


 a 
R2 := p ⋅ ( f − e) − R1 + P
R1 = 6275⋅ lbf
R2 = 7525⋅ lbf
6.
Define the range for x
7.
For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
x := 0 ⋅ m , 0.002⋅ L .. L
S ( x , z) := if ( x ≥ z , 1 , 0 )
8.
Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the
effect of the singularity functions.
V ( x) := R1⋅ S ( x , 0 ⋅ m) − p ⋅ S ( x , e) ⋅ ( x − e) + p ⋅ S ( x , f ) ⋅ ( x − f ) + R2⋅ S ( x , a ) − P⋅ S ( x , b )
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-39a-2
p
2
2
⋅ S ( x , e) ⋅ ( x − e) + ⋅ S ( x , f ) ⋅ ( x − f ) ...
2
2
+ R2⋅ S ( x , a ) ⋅ ( x − a) − P⋅ S ( x , b ) ⋅ ( x − b )
M ( x) := R1⋅ S ( x , 0 ⋅ m) ⋅ x −
9.
p
Plot the shear and moment diagrams.
SHEAR DIAGRAM
MOMENT DIAGRAM
10000
30000
5000
20000
V ( x)
lbf
M ( x)
0
in⋅ lbf
10000
− 5000
− 10000
0
0
5
10
15
− 10000
20
0
5
10
x
x
in
in
15
20
FIGURE 3-39aB
Shear and Moment Diagrams for Problem 3-39a
9.
Determine the maximum shear and maximum moment from inspection of the diagrams.
Maximum shear:
Vmax := V ( f )
Vmax = 6525⋅ lbf
Maximum moment occurs where V is zero, which is x = c:
c−e
R1
=
f −c
R2 − P
Mmax := M ( c)
c :=
f ⋅ R1 + e ⋅ R2 − e ⋅ P
R1 + R2 − P
c = 7.875⋅ in
Mmax = 29728⋅ in⋅ lbf
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-40a-1
PROBLEM 3-40a
Statement:
A beam is supported and loaded as shown in Figure P3-18. Find the reactions, maximum shear, and
maximum moment for the data given in row a from Table P3-2.
a
Given:
Distance to gear 2
L := 20⋅ in
Distance to gear 1
a := 16⋅ in
Distance to RH bearing
b := 18⋅ in
Concentrated load at gear 2
P2 := 1000⋅ lbf
Concentrated load at gear 1
P1 := 0.4⋅ P2
P1
P2
R2
R1
b
L
FIGURE 3-40a
Solution:
1.
See Figure 3-40 and Mathcad file P0340a.
Free Body Diagram for Problem 3-40
From inspection of Figure 3-40, write the load function equation
q(x) = R1<x - 0>-1 - P1<x - a>-1 + R2<x - b>-1 - P2<x - L>-1
2.
Integrate this equation from -∞ to x to obtain shear, V(x)
V(x) = R1<x - 0>0 - P1<x - a>0 + R2<x - b>0 - P2<x - L>0
3.
Integrate this equation from -∞ to x to obtain moment, M(x)
M(x) = R1<x - 0>1 - P1<x - a>1 + R2<x - b>1 - P<x - L>1
4.
Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = L, where
both are zero.
At x = L+, V = M = 0
V = R1 − P1 + R2 − P2 = 0
M = R1⋅ L − P1⋅ ( L − a ) + R2⋅ ( b − a ) = 0
R1 := P1⋅  1 −


 + P2⋅  1 −
b

a
L

b
R2 := P1 + P2 − R1
R1 = −67⋅ lbf
R2 = 1467⋅ lbf
5.
Define the range for x
6.
For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
x := 0 ⋅ m , 0.002⋅ L .. L
S ( x , z) := if ( x ≥ z , 1 , 0 )
7.
Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the
effect of the singularity functions.
V ( x) := R1⋅ S ( x , 0 ⋅ in) − P1⋅ S ( x , a ) + R2⋅ S ( x , b ) − P2⋅ S ( x , L)
M ( x) := R1⋅ S ( x , 0 ⋅ mm) ⋅ ( x − 0 ⋅ mm) − P1⋅ S ( x , a ) ⋅ ( x − a ) ...
+ R2⋅ S ( x , b ) ⋅ ( x − b) − P2⋅ S ( x , L) ⋅ ( x − L)
8.
Plot the shear and moment diagrams.
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-40a-2
SHEAR DIAGRAM
MOMENT DIAGRAM
1000
0
500
− 1000
V ( x)
M ( x)
lbf
in⋅ lbf
0
− 500
− 2000
0
5
10
15
20
− 3000
0
5
10
x
x
in
in
15
20
FIGURE 3-40aB
Shear and Moment Diagrams for Problem 3-40a
9.
Determine the maximum shear and maximum moment from inspection of the diagrams.
Maximum shear:
Vmax := V ( b )
Vmax = 1000⋅ lbf
Maximum moment occurs where V is zero, which is x = b:
Mmax := M ( b )
Mmax = 2000⋅ in⋅ lbf
© 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,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-41a-1
PROBLEM 3-41a
Statement:
Input data:
A beam is supported and loaded as shown in Figure P3-18. Write a computer program or equation
solver model to find the reactions and calculate and plot the loading, shear, and moment functions.
Test the program with the data given in row a from Table P3-2.
a
Enter data in highlighted areas
Distance to gear 2
L := 20⋅ in
Distance to gear 1
a := 16⋅ in
Distance to RH bearing
b := 18⋅ in
Concentrated load at gear 2
P2 := 1000⋅ lbf
Concentrated load at gear 1
P1 := 0.4⋅ P2
P1
P2
R2
R1
b
L
FIGURE 3-41aA
Solution:
1.
See Figure 3-41 and Mathcad file P0341a.
Free Body Diagram for Problem 3-41
From inspection of Figure 3-41, write the load function equation
q(x) = R1<x - 0>-1 - P1<x - a>-1 + R2<x - b>-1 - P2<x - L>-1
2.
Integrate this equation from -∞ to x to obtain shear, V(x)
V(x) = R1<x - 0>0 - P1<x - a>0 + R2<x - b>0 - P2<x - L>0
3.
Integrate this equation from -∞ to x to obtain moment, M(x)
M(x) = R1<x - 0>1 - P1<x - a>1 + R2<x - b>1 - P<x - L>1
4.
Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = L, where
both are zero.
At x = L+, V = M = 0
V = R1 − P1 + R2 − P2 = 0
M = R1⋅ L − P1⋅ ( L − a ) + R2⋅ ( b − a ) = 0
R1 := P1⋅  1 −


 + P2⋅  1 −
b

a
L

b
R2 := P1 + P2 − R1
R1 = −67⋅ lbf
R2 = 1467⋅ lbf
5.
Define the range for x
6.
For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
x := 0 ⋅ m , 0.002⋅ L .. L
S ( x , z) := if ( x ≥ z , 1 , 0 )
7.
Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the
effect of the singularity functions.
V ( z) := R1⋅ S ( z , 0 ⋅ in) − P1⋅ S ( z , a) + R2⋅ S ( z , b) − P2⋅ S ( z , L)
M ( z) := R1⋅ S ( z , 0 ⋅ mm) ⋅ ( z − 0 ⋅ mm ) − P1⋅ S ( z , a) ⋅ ( z − a ) ...
+ R2⋅ S ( z , b) ⋅ ( z − b ) − P2⋅ S ( z , L) ⋅ ( z − L)
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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,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
8.
3-41a-2
Plot the shear and moment diagrams.
SHEAR DIAGRAM
MOMENT DIAGRAM
1000
0
500
− 1000
V ( x)
M ( x)
lbf
in⋅ lbf
0
− 500
− 2000
0
5
10
15
20
− 3000
0
5
10
x
x
in
in
15
20
FIGURE 3-41aB
Shear and Moment Diagrams for Problem 3-41a
9.
Determine the maximum shear and maximum moment from inspection of the diagrams.
Maximum shear:
Vmax := V ( b )
Vmax = 1000⋅ lbf
Maximum moment occurs where V is zero, which is x = b:
Mmax := M ( b )
Mmax = 2000⋅ in⋅ lbf
© 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,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
PROBLEM 3-42
3-42-1
_____
Statement:
A 1000 kg speedboat reaches a speed of 16 kph at the instant it takes up the slack in a 100 m-long
tow rope attached to a surfboard carrying a 100 kg passenger. If the rope has k = 5 N/m, what is
the dynamic force exerted on the surfboard?
Given:
Mass of speedboat
ms := 1000⋅ kg
Speed of boat
vi := 16⋅ kph
Mass of passenger
mp := 100⋅ kg
Rope stiffness
k := 5 ⋅ N ⋅ m
−1
Assumptions: 1. The water does not influence the dynamic force.
2. An impact model can be used to estimate the dynamic force.
Solution:
1.
See Mathcad file P0342.
For the impact model, the passenger is the "struck" mass and the speedboat is the "striking mass". Thus, from
equation 3.15, the energy correction factor is:
1
η :=
1+
2.
mp
η = 0.97
3 ⋅ ms
Use equation 3.11 to estimate the dynamic force on the surfboard/passenger.
Fi := vi⋅ η⋅ ms⋅ k
Fi = 309⋅ 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,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-43-1
PROBLEM 3-43
Statement:
Figure P3-19 shows an oil-field pump jack. For the position shown, draw free-body diagrams of the
crank (2), connecting rod (3) and walking beam (4) using variable names similar to those used in
Case Studies 1A and 2A. Assume that the crank turns slowly enough that accelerations can be
ignored. Include the weight acting at the CG of the walking beam and the crank but not the
connecting rod.
Assumptions: 1. A two-dimensional model is adequate.
2. Inertia forces may be ignored.
Solution:
1.
See Mathcad file P0343.
Isolate each of the elements to be analyzed, starting with the walking beam, since the external forces on it are
known. Place the known force, Fcable, at the point P and the known weight at the CG. Assume the forces at the
interfaces O4 and B to be positive. The position vectors R14, R34, and Rp will be known as will the angle, θ3,that
the connecting rod makes with the horizontal axis.
F34
R34
y
R 14
RP
head end
P
F
B
x
14y
4
counterweight
O4
F cable
2.
θ3
F
14x
W4
F43
The connecting rod is a two-force member with
the forces acting at the interfaces A and B along
the line joining points A and B. The assumption
made in step 1 is that these are compressive
forces on link 3.
y
B
R 43
3
θ3
x
R23
A
F 23
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obtained from
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P0343.xmcd
mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3.
3-43-2
The crank is acted on by forces at A and O2, its weight at its CG, and a torque which we will assume to be
positive (CCW). As in step 1, assume that the unknown reaction force at O2 is positive.
F32
y
F
A
12y
θ3
x
2
O2
F
12x
T2
W2
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obtained from
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P0343.xmcd
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-44-1
PROBLEM 3-44
Statement:
For the pump jack of Problem 3-43 and the data of Table P3-3, determine the pin forces on the
walking beam, connecting rod, and crank and the reaction torque on the crank.
Given:
R12 := 13.2⋅ in
θ12 := 135⋅ deg
R14 := 79.22⋅ in
θ14 := 196⋅ deg
R32 := 0.80⋅ in
θ32 := 45⋅ deg
R34 := 32.00⋅ in
θ34 := 169⋅ deg
Fcable := 2970⋅ lbf
W2 := 598⋅ lbf
θ3 := 98.5⋅ deg
Solution:
1.
W4 := 2706⋅ lbf
RP := 124.44⋅ in θP := 185⋅ deg
See Mathcad files P0343 and P0344.
Draw free-body diagrams of each element (see Problem 3-43).
F34
R34
y
R 14
θ3
RP
head end
F
P
B
x
14y
4
counterweight
O4
F
14x
W4
F cable
F43
y
B
F32
y
F
R 43
A
12y
3
θ3
θ3
x
x
2
O2
F
12x
R23
T2
W2
A
F 23
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
obtained from
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P0344.xmcd
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
2. Calculate the x- and y-components of the position vectors.
3.
4.
3-44-2
R12x := R12⋅ cos( θ12)
R12x = −9.334⋅ in
R12y := R12⋅ sin( θ12)
R12y = 9.334⋅ in
R14x := R14⋅ cos( θ14)
R14x = −76.151⋅ in
R14y := R14⋅ sin( θ14)
R14y = −21.836⋅ in
R32x := R32⋅ cos( θ32)
R32x = 0.566⋅ in
R32y := R32⋅ sin( θ32)
R32y = 0.566⋅ in
R34x := R34⋅ cos( θ34)
R34x = −31.412⋅ in
R34y := R34⋅ sin( θ34)
R34y = 6.106⋅ in
RPx := RP⋅ cos( θP)
RPx = −123.966 ⋅ in
RPy := RP⋅ sin( θP)
RPy = −10.846⋅ in
Write equations 3(b) for link 4, the walking beam.
Σ Fx:
F14x + F34x = 0
(1)
Σ Fy:
−Fcable + F14y + F34y − W4 = 0
(2)
Σ Mz:
Rpx⋅ Fcable + ( R14x⋅ F14y − R14y⋅ F14x) + ( R34x⋅ F34y − R34y⋅ F34x) = 0
(3)
The direction (but not the sense) of F34 is known so write the equation that relates the x- and y-components of
this force.
F34y − F34x⋅ tan( θ3) = 0
5.
(4)
There are four unknowns in the four equations above. Solving for F14x, F14y, F34x, and F34y,
0
1
0 
 1
 0
1
0
1 


A :=  −R14y R14x −R34y R34x 
 in
in
in 
in


0 −tan( θ3)
1 
 0
F14x = 2446⋅ lbf
6.
7.
8.
F14y = −10687⋅ lbf
0


F
+
W
4
 cable


lbf
B := 

 −RPx ⋅ Fcable 
 in⋅ lbf 


0
F34x = −2446⋅ lbf
 F14x 
F 
 14y  := A − 1⋅ B⋅ lbf
 F34x 


 F34y 
F34y = 16363⋅ lbf
From Newton's thrid law and, since the connecting rod (3) is a two-force member
F43x := −F34x
F43x = 2446⋅ lbf
F43y := −F34y
F43y = −16363⋅ lbf
F23x := −F43x
F23x = −2446⋅ lbf
F23y := −F43y
F23y = 16363⋅ lbf
Write equations 3(b) for link 2, the crank.
Σ Fx:
F12x + F32x = 0
(5)
Σ Fy:
F12y + F32y − W2 = 0
(6)
Σ Mz:
T2 + ( R12x⋅ F12y − R12y⋅ F12x) + ( R32x⋅ F32y − R32y⋅ F32x) = 0
(7)
There are three unknowns in the three equations above. Solving for F12x, F12y, and T2, since
F32x := −F23x
F32x = 2446⋅ lbf
F32y := −F23y
F32y = −16363⋅ lbf
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
obtained from
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P0344.xmcd
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Pearson Education, Inc., Upper Saddle River, NJ 07458.
MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-44-3
−F32x


lbf




W2 − F32y
B := 

lbf


 −( R32x⋅ F32y − R32y⋅ F32x) 


in⋅ lbf


0 0
 1


0
1 0

A :=
 −R12y R12x 
1
 in
in


F12x = −2446
lbf
T2 = 146128
in-lbf
F12y = 16961
 F12x 


−1
 F12y  := A ⋅ B
 T 
 2 
lbf
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
obtained from
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-45-1
PROBLEM 3-45
Statement:
Figure P3-20 shows an aircraft overhead bin mechanism in end view. For the position shown, draw
free-body diagrams of links 2 and 4 and the door (3) using variable names similar to those used in
Case Studies 1A and 2A. There are stops that prevent further clockwise motion of link 2 (and the
identical link behind it at the other end of the door) resulting in horizontal forces being applied to
the door at points A. Assume that the mechanism is symmetrical so that each set of links 2 and 4
carry one half of the door weight. Ignore the weight of links 2 and 4 as they are negligible.
Assumptions: 1. A two-dimensional model is adequate.
2. Inertia forces may be ignored as the mechanism is at rest against stops.
Solution:
1.
See Mathcad file P0345.
Isolate each of the elements to be analyzed, starting with the door. Place the force, Fstop, at the point A and the
known weight at the CG. Assume the forces in links 2 and 4 to be positive (tensile). The position vectors R43
and R23 will be known as will the angles θ2 and θ4 that links 2 and 4 make with the horizontal axis.
F 23
θ4
F 43
θ2
y
F stop
R43
R23
A
3
B
x
W3
2
2.
F 12
Links 2 and 4 are two-force members with the
forces acting at the pinned ends along the line
joining the pin centers. The assumption made in
step 1 is that these are tensile forces on links 2
and 4.
θ2
y
O2
R12
2
x
θ4
y
R32
F 14
A
O4
R14
x
4
R34
B
F 34
F 32
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-46-1
PROBLEM 3-46
Statement:
For the overhead bin mechanism of Problem 3-45 and the data of Table P3-4, determine the pin
forces on the door (3), and links 2 & 4 and the reaction force on each of the two stops.
Given:
R23 := 180.0⋅ mm θ23 := 160.345⋅ deg
W3 := 45⋅ N
Solution:
1.
θ43 := 27.862⋅ deg
R43 := 180.0⋅ mm
θ2 := 85.879⋅ deg
θ4 := 172.352 ⋅ deg
See Mathcad files P0345 and P0346.
Draw free-body diagrams of each element (see Problem 3-45).
F 12
θ2
y
O2
θ4
y
F 14
R12
2
O4
x
x
R14
4
R34
F 34
B
R32
A
F 23
θ4
F 43
θ2
F 32
y
F stop
A
R23
R43
3
B
x
W3
2
2.
3.
Calculate the x- and y-components of the position vectors on the door (3).
R23x := R23⋅ cos( θ23)
R23x = −169.512 ⋅ mm
R23y := R23⋅ sin( θ23)
R23y = 60.544⋅ mm
R43x := R43⋅ cos( θ43)
R43x = 159.134 ⋅ mm
R43y := R43⋅ sin( θ43)
R43y = 84.122⋅ mm
Write equations 3(b) for link 3, the door.
Σ Fx:
Fstop + F23x + F43x = 0
(1)
Σ
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
F23y + F43y − 0.5⋅ W3 = 0
Σ Fy:
Σ Mz:
4.
5.
3-46-2
(2)
−R23x⋅ Fstop + ( R23x⋅ F23y − R23y⋅ F23x) + ( R43x⋅ F43y − R43y⋅ F43x) = 0
(3)
The directions (but not the sense) of F 23 and F43 are known so write the equations that relates the x- and
y-components of these forces.
F23y − F23x⋅ tan( θ2) = 0
(4)
F43y − F43x⋅ tan( θ4) = 0
(5)
There are five unknowns in the five equations above. Solving for F23x, F23y, F43x, F43y, and Fstop:
0
1
0
1 
 1


1
0
1
0 
 0
 −R23y R23x −R43y R43x −R23x 
A := 

mm
mm
mm
mm
mm


0
0
0 
 −tan( θ2) 1
 0
0 −tan( θ4)
1
0 

F23x = 1.49⋅ N
F23y = 20.63⋅ N
 0 
 0.5⋅ W3 


N 

B :=
 0 
 0 


 0 
F43x = −13.96⋅ N
 F23x 


 F23y 
 F43x  := A − 1⋅ B⋅ N


 F43y 
F 
 stop 
F43y = 1.87⋅ N
The pin forces at A and B are:
F23 :=
2
F23x + F23y
The force on each stop is:
6.
2
F23 = 20.68⋅ N
F43 :=
2
F43x + F43y
2
F43 = 14.08⋅ N
Fstop = 12.47⋅ N
From Newton's thrid law and, since links 2 and 4 are two-force members
F34x := −F43x
F34x = 13.96⋅ N
F34y := −F43y
F34y = −1.87⋅ N
F32x := −F23x
F32x = −1.49⋅ N
F32y := −F23y
F32y = −20.63⋅ N
The pin forces at O2 and O4 are numerically equal to those at A and B, respectively.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-47-1
PROBLEM 3-47
Statement:
A particular automobile front suspension consists of two A-arms , the wheel (with tire), a coil
spring, and a shock absorber (damper). The effective stiffness of the suspension (called the "ride
rate") is a function of the coil spring stiffness and the tire stiffness. The A-arms are designed to
give the wheel a nearly vertical displacement as the tire rides over bumps in the road. The entire
assembly can be modeled as a spring-mass-damper system as shown in Figure 3-15(b). If the
sprung mass (mass of the portion of the vehicle supported by the suspension system) weighs 675
lb, determine the ride rate that is required to achieve an undamped natural frequency of 1.4 Hz.
What is the static deflection of the suspension for the calculated ride rate?
Units:
Hz := 2 ⋅ π⋅ rad⋅ sec
Given:
Sprung mass
Solution:
See Figure 3-15(b) and Mathcad file P0347.
−1
Ws := 675⋅ lbf
Ws
Natural frequency ωn := 1.4⋅ Hz
2
1.
Calculate the sprung mass Ms :=
2.
Using equation 3.4, calculate the required ride rate
Ride rate
3.
−1
Ms = 1.748 lbf ⋅ sec ⋅ in
g
2
k := ωn ⋅ Ms
k = 135.28
lbf
in
Calculate the static deflection using equation 3.5
Static deflection
δ :=
Ws
k
δ = 4.99in
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-48-1
PROBLEM 3-48
Statement:
The independent suspension system of Problem 3-47 has an unsprung weight (the weight of the
axle, wheel, A-arms, etc.) of 106 lb. Calculate the natural frequency (hop resonance) of the
unsprung mass if the sum of the tire and coil spring stiffnesses is 1100 lb/in.
Units:
Hz := 2 ⋅ π⋅ rad⋅ sec
Given:
Unsprung mass
Solution:
−1
Wu := 106⋅ lbf
Stiffness
k := 1100⋅
lbf
in
See Figure 3-15(b) and Mathcad file P0348.
Wu
2
1.
Calculate the unsprung mass Mu :=
2.
Using equation 3.4, calculate the natural frequency
Natural frequency
−1
Mu = 0.275 lbf ⋅ sec ⋅ in
g
ωn :=
k
Mu
ωn = 10.1 Hz
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-49-1
PROBLEM 3-49
Statement:
The independent suspension system of Problem 3-47 has a sprung weight of 675 lb and a ride rate
of 135 lb/in. Calculate the damped natural frequency of the sprung mass if the damping coefficient
of the shock absorber is a constant 12 lb-sec/in.
Units:
Hz := 2 ⋅ π⋅ rad⋅ sec
Given:
Sprung mass
−1
Ws := 675⋅ lbf
Damping coefficient
Solution:
d := 12⋅
Ride rate
k := 135⋅
lbf
in
lbf ⋅ sec
in
See Figure 3-15(b) and Mathcad file P0349.
Ws
2
1.
Calculate the sprung mass Ms :=
2.
Using equation 3.7, calculate the damped natural frequency
g
Ms = 1.748lbf ⋅ sec ⋅ in
Damped natural frequency
ωd :=
k
Ms
−1
− 


 2⋅ Ms 
d
2
ωd = 1.29Hz
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-50-1
PROBLEM 3-50_______________________________________________________
Statement:
Figure P3-22 shows a powder compaction mechanism. For the position shown, draw free-body
diagrams of the input arm (2), connecting rod (3) and compacting ram (4) using variable names
similar to those used in Case Studies 1A and 2A. Assume that the input arm turns slowly
enough that accelerations can be ignored. Ignore the weights of the arm, connecting rod, and
compacting ram. Neglect friction.
Assumptions: 1. A two-dimensional model is adequate.
2. Inertia forces may be ignored.
3. The reactions at slider bearings E and F can be modeled as concentrated forces acting
horizontally at the center of each bearing.
See Mathcad file P0350.
Solution:
1.
y
Isolate each of the elements to be analyzed, starting with the
compacting rod, since the external force on it is known. Place the
known force, Fcom , at the point P. The position vectors R14E,
F14E
R14F, and R p will be known as will the angle, q3,that the
compacting ram makes with the vertical axis.
2.
3.
R14E
E
R34
D
The connecting rod is a two-force member with the forces
acting at the interfaces B and D along the line joining points B
and D. The assumption made in step 1 is that these are tensile
forces on link 3.
The input arm is acted on by forces at A, B, and C. Assume that the
unknown reaction force at A is positive.
F14F
x
F34
3
F43
F
R14F
RP
P
Fcom
D
Compacting Ram (4)
y
R43
x
C
R23
B
F23
Connecting Rod (3)
y
Fin
F32
x
Rin
B
R32
F12y
A
R12
F12x
Input Arm (2)
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-51-1
PROBLEM 3-51______________________________________________________
Statement:
For the compaction mechanism of Problem 3-50 and the data of Table P3-5, determine the pin forces
on the compacting ram, connecting rod, and input arm.
Given:
R12 := 148.4⋅ mm θ12 := −45⋅ deg
R14E := 57.0⋅ mm θ14E := 90⋅ deg
R14F := 62.9⋅ mm θ14F := 270⋅ deg
R34 := 32.00⋅ in
R23 := 87.6⋅ mm
θ23 := 254.36⋅ deg
R32 := 42.9⋅ mm
θ32 := 74.36⋅ deg
Solution:
1.
θ43 := 74.36⋅ deg
R43 := 87.6⋅ mm
θ34 := 90⋅ deg
R34 := 15.0⋅ mm
Rin := 152.6⋅ mm θin := 225⋅ deg
Fcom := 100⋅ N
θ34 := −105.64⋅ deg
RP := 105.0⋅ mm θP := 270⋅ deg
θ3 := 254.36⋅ deg
See Mathcad files P0350 and P0351.
Draw free-body diagrams of each element (see Problem
3-50).
y
F43
R14E
E
F14E
R34
D
D
x
F34
y
F14F
R43
R14F
F
θ3
RP
P
x
Fcom
R23
Compacting Ram (4)
B
C
F23
y
Connecting Rod (3)
Fin
F32
x
Rin
B
R32
Input Arm (2)
F12y
A
R12
F12x
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
2.
3.
4.
3-51-2
Calculate the x- and y-components of the position vectors.
R12x := R12⋅ cos( θ12)
R12x = 104.935 ⋅ mm
R12y := R12⋅ sin( θ12)
R12y = −104.935 ⋅ mm
R14Ex := R14E⋅ cos( θ14E)
R14Ex = 0 ⋅ mm
R14Ey := R14E⋅ sin( θ14E)
R14Ey = 57.000⋅ mm
R14Fx := R14F ⋅ cos( θ14F )
R14Fx = −0.000⋅ mm
R14Fy := R14F ⋅ sin( θ14F )
R14Fy = −62.900⋅ mm
R23x := R23⋅ cos( θ23)
R23x = −23.616⋅ mm
R23y := R23⋅ sin( θ23)
R23y = −84.357⋅ mm
R32x := R32⋅ cos( θ32)
R32x = 11.566⋅ mm
R32y := R32⋅ sin( θ32)
R32y = 41.312⋅ mm
R34x := R34⋅ cos( θ34)
R34x = 0.000⋅ mm
R34y := R34⋅ sin( θ34)
R34y = 15.000⋅ mm
R43x := R43⋅ cos( θ43)
R43x = 23.616⋅ mm
R43y := R43⋅ sin( θ43)
R43y = 84.357⋅ mm
RPx := RP⋅ cos( θP)
RPx = −0.000⋅ mm
RPy := RP⋅ sin( θP)
RPy = −105.000 ⋅ mm
Rinx := Rin⋅ cos( θin)
Rinx = −107.904 ⋅ mm
Riny := Rin⋅ sin( θin)
Riny = −107.904 ⋅ mm
Write equations 3(b) for link 4, the compacting ram.
Σ Fx:
F14E + F14F + F34x = 0
(1)
Σ Fy:
Fcom + F34y = 0
(2)
Σ Mz:
(−R14Ey⋅ F14E) + (−R14Fy⋅ F14F ) + (R34x⋅ F34y − R34y⋅ F34x) = 0
(3)
The direction (but not the sense) of F34 is known so write the equation that relates the x- and y-components of
this force.
F34y − F34x⋅ tan( θ3) = 0
5.
(4)
There are four unknowns in the four equations above. Solving for F14x, F14y, F34x, and F34y,
1
1
0 
 1
 0
0
0
1 


A :=  −R14Ey −R14Fy −R34y R34x 
 mm
mm
mm
mm 


0
−tan( θ3)
1 
 0
F14E = 18.2⋅ N
6.
7.
F14F = 9.8⋅ N
 0 
 F 
 − com 
B :=  N 
 0 


 0 
F34x = −28.0⋅ N
 F14E 
F 
 14F  := A − 1⋅ B⋅ N
 F34x 


 F34y 
F34y = −100.0⋅ N
From Newton's thrid law and, since the connecting rod (3) is a two-force member
F43x := −F34x
F43x = 28.0⋅ N
F43y := −F34y
F43y = 100.0⋅ N
F23x := −F43x
F23x = −28.0⋅ N
F23y := −F43y
F23y = −100.0⋅ N
Write equations 3(b) for link 2, the input arm.
Σ Fx:
F12x + F32x + Finx = 0
(5)
Σ Fy:
F12y + F32y + Finy = 0
(6)
Σ Mz:
(R12x⋅ F12y − R12y⋅ F12x) + (R32x⋅ F32y − R32y⋅ F32x) + (Rinx⋅ Finy − Riny⋅ Finx) = 0
(7)
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
8.
3-51-3
The direction (but not the sense) of Fin is known so write the equation that relates the x- and
y-components of this force.
Finy − Finx⋅ tan( θin) = 0
9.
(8)
There are four unknowns in the four equations above. Solving for F12x, F12y, Finx, and Finy, since
F32x := −F23x
F32x = 28⋅ N
0
1
0 
 1
 0
1
0
1 


−Riny
Rinx 
A :=  −R12y R12x
 mm mm
mm
mm 


0 −tan( θin) 1 
 0
F12x = 36.0⋅ N
F12 :=
F12y = −36.0⋅ N
 F12x2 + F12y2


F12 = 51⋅ N
F32y = 100⋅ N
F32y := −F23y
−F32x




N



−F32y



B :=
N


 −( R32x⋅ F32y − R32y⋅ F32x) 


N ⋅ mm


0


Finx = −64.0⋅ N
Fin :=
 F12x 
F 
 12y  := A − 1⋅ B⋅ N
 Finx 


 Finy 
Finx = −64.0⋅ N
 Finx2 + Finy2


Fin = 91⋅ N
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-52-1
PROBLEM 3-52
Statement:
Figure P3-23 shows a drag link slider crank mechanism. For the position shown, draw free-body
diagrams of links 2 through 6 using variable names similar to those used in Case Studies 1A and
2A. Assume that the crank turns slowly enough that accelerations can be ignored. Ignore the
weights of the links and any friction forces or torques.
Assumptions: 1. A two-dimensional model is adequate.
2. Inertia forces may be ignored.
3. Links 4 and 6 are three-force bodies.
Solution:
1.
See Figure P3-23 and Mathcad file P0352.
Isolate each of the elements to be analyzed,
starting with the slider, link 6, since the external
forces on it are known. Place the known force,
FP, at the point P. This is a three-force member
y
F56
θ5
so the forces are coincident at point D and there
is no turning moiment on the link. The angle,
θ5,that link 5 makes with the horizontal axis is
known.
2.
D
FP
F16
Link 5 is a two-force member with the forces
acting at the interfaces C and D along the line
joining points C and D. The assumption made in
step 1 is that these are compressive forces on link
5.
F45
P
x
Slider block 6
C
y
x
θ5
R45
D
R65
F65
Link 5
3.
Link 4 is a three-force body with the three forces meeting at a point. The position vectors R 14, R34, and
R54 will be known as will the angles,θ 3 and θ5,that links 3 and 5, respectively, make with the horizontal
axis.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-52-2
C
y
R34
F54
E
R54
x
F14y
B
O4
F14x
R14
F34
F43
Link 4
y
B
4.
Link 3 is a two-force member with the forces acting at the
interfaces A and B along the line joining points A and B.
x
R23
5.
R43
The crank is acted on by forces at A and O2, and a torque
which we will assume to be positive (CCW). As in step 1,
assume that the unknown reaction force at O2 is positive.
A
F23
Link 3
F
y
12y
T
F12x
R32
O2
x
R12
A
F32
Link 2
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-53-1
PROBLEM 3-53
Statement:
For the drag link slider crank mechanism of Problem 3-52 and the data of Table P3-6, determine the
pin forces on the slider, connecting rods, and crank and the reaction torque on the crank.
Given:
R12 := 63.5⋅ mm
θ12 := 45.38⋅ deg
R14 := 93.6⋅ mm
θ14 := −55.89⋅ deg
R23 := 63.5⋅ mm
θ23 := 267.8⋅ deg
R32 := 63.5⋅ mm
θ32 := 225.38⋅ deg
R34 := 103.5⋅ mm θ34 := 202.68⋅ deg
θ43 := 87.80⋅ deg
R45 := 190.5⋅ mm θ45 := 156.65⋅ deg
R54 := 103.5⋅ mm θ54 := 45.34⋅ deg
R65 := 190.5⋅ mm θ65 := −23.35⋅ deg
R43 := 63.5⋅ mm
θ5 := 156.65deg
FP := 85⋅ N
Solution:
1.
θ3 := 87.80⋅ deg
See Mathcad files P0352 and P0353.
Draw free-body diagrams of each element (see
Problem 3-52).
y
F56
θ5
Slider block 6
P
FP
x
D
F16
F45
C
y
x
θ5
R45
D
Link 5
2.
R65
F65
Calculate the x- and y-components of the position vectors.
R12x := R12⋅ cos( θ12)
R12x = 44.602⋅ mm
R12y := R12⋅ sin( θ12)
R12y = 45.198⋅ mm
R14x := R14⋅ cos( θ14)
R14x = 52.489⋅ mm
R14y := R14⋅ sin( θ14)
R14y = −77.497⋅ mm
R23x := R23⋅ cos( θ23)
R23x = −2.438⋅ mm
R23y := R23⋅ sin( θ23)
R23y = −63.453⋅ mm
R32x := R32⋅ cos( θ32)
R32x = −44.602⋅ mm
R32y := R32⋅ sin( θ32)
R32y = −45.198⋅ mm
R34x := R34⋅ cos( θ34)
R34x = −95.497⋅ mm
R34y := R34⋅ sin( θ34)
R34y = −39.908⋅ mm
R43x := R43⋅ cos( θ43)
R43x = 2.438⋅ mm
R43y := R43⋅ sin( θ43)
R43y = 63.453⋅ mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-53-2
R45x := R45⋅ cos( θ45)
R45x = −174.898 ⋅ mm
R45y := R45⋅ sin( θ45)
R45y = 75.504⋅ mm
R54x := R54⋅ cos( θ54)
R54x = 72.75⋅ mm
R54y := R54⋅ sin( θ54)
R54y = 73.619⋅ mm
R65x := R65⋅ cos( θ65)
R65x = 174.898 ⋅ mm
R65y := R65⋅ sin( θ65)
R65y = −75.504⋅ mm
C
F43
y
y
R34
B
F54
E
R43
x
R54
x
F14y
B
O4
F14x
R23
A
R14
F34
Link 4
F23
Link 3
F
y
12y
T
F12x
R32
O2
x
R12
A
F32
Link 2
3.
4.
Write equations 3(b) for link 5, the slider.
Σ Fx:
F56x − FP = 0
(1)
Σ Fy:
F16 + F56y = 0
(2)
The direction (but not the sense) of F56 is known so write the equation that relates the x- and y-components of
this force.
F56y − F56x⋅ tan( θ5) = 0
5.
There are three unknowns in the three equations above. Solving for F56x, F56y, and F16,
(3)
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
 FP 
 
N 
B := 
 0 
 0 
 
0 0
 1
0
1 1
A := 


 −tan( θ5) 1 0 
F56x = 85.0⋅ N
6.
3-53-3
F56y = −36.7⋅ N
 F56x 


−1
 F56y  := A ⋅ B⋅ N
F 
 16 
F16 = 36.7⋅ N
From Newton's thrid law and, since the connecting rod (5) is a two-force member
F65x := −F56x
F65x = −85⋅ N
F65y := −F56y
F65y = 36.7⋅ N
F45x := −F65x
F45x = 85⋅ N
F45y := −F65y
F45y = −36.7⋅ N
F54x = −85⋅ N
F54y := −F45y
F54y = 36.7⋅ N
and, for link 4
F54x := −F45x
7.
8.
Write equations 3(b) for link 4, the rocker.
Σ Fx:
F34x + F54x + F14x = 0
(4)
Σ Fy:
F34y + F54y + F14y = 0
(5)
Σ Mz:
(R14x⋅ F14y − R14y⋅ F14x) + (R34x⋅ F34y − R34y⋅ F34x) + (R54x⋅ F54y − R54y⋅ F54x) = 0
The direction (but not the sense) of F34 is known so write the equation that relates the x- and y-components of
this force.
F34y − F34x⋅ tan( θ3) = 0
9.
(7)
There are four unknowns in the four equations above. Solving for F34x, F34y, F14x, and F14y,
0
1
0 
 1
 0
1
0
1 


A :=  −R34y R34x −R14y R14x 
 mm
mm
mm
mm 


0
0 
 −tan( θ3) 1
F34x = 3.5⋅ N
F34y = 90.9⋅ N
−F54x




N



−F54y



B :=
N


 −( R54x⋅ F54y − R54y⋅ F54x) 


N ⋅ mm


0


F14x = 81.5⋅ N
 F34x 
F 
 34y  := A − 1⋅ B⋅ N
 F14x 


 F14y 
F14y = −127.6⋅ N
10. From Newton's thrid law and, since the connecting rod (3) is a two-force member
F43x := −F34x
F43x = −3.5⋅ N
F43y := −F34y
F43y = −90.9⋅ N
F23x := −F43x
F23x = 3 ⋅ N
F23y := −F43y
F23y = 90.9⋅ N
F32x = −3.5⋅ N
F32y := −F23y
F32y = −90.9⋅ N
and, for link 2
F32x := −F23x
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3-53-4
11. Write equations 3(b) for link 2, the crank.
Σ Fx:
F12x + F32x = 0
(8)
Σ Fy:
F12y + F32y = 0
(9)
Σ Mz:
T2 + ( R12x⋅ F12y − R12y⋅ F12x) + ( R32x⋅ F32y − R32y⋅ F32x) = 0
(10)
12. There are three unknowns in the three equations above. Solving for F12x, F12y, and T2
0 0
 1


0
1 0

A :=
 −R12y R12x 
 mm mm 1 


F12x = 3.5 N
−F32x


N




−F32y
B := 

N


 −( R32x⋅ F32y − R32y⋅ F32x) 


N ⋅ mm


F12y = 90.9
N
 F12x 


−1
 F12y  := A ⋅ B
 T 
 2 
T2 = −7796
N*mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-1a-1
PROBLEM 4-1a
Statement:
A differential stress element has a set of applied stresses on it as indicated in each row of Table
P4-1. For row a, draw the stress element showing the applied stresses, find the principal stresses
and maximum shear stress using Mohr's circle diagram, and draw the rotated stress element
showing the principal stresses.
Given:
σx := 1000
σy := 0
σz := 0
τxy := 500
τyz := 0
τzx := 0
Solution:
See Figure 4-1a and Mathcad file P0401a.
500
y
1. Draw the stress element, indicating the x and y axes.
1000
x
2. Draw the Mohr's circle axes, indicating the τ and σ
axes with CW up and CCW down.
3. Plot the positive x-face point, which is (+1000, -500),
and label it with an "x."
FIGURE 4-1aA
4. Plot the positive y-face point, which is (0, +500), and
label it with a "y."
Stress Element for Problem 4-1a
5. Draw a straight line from point x to point y. Using the point where this line intersects the σ-axis as the center of
the Mohr circle, draw a circle that goes through points x and y.
6. The center of the circle will be at
σc :=
σx + σy
σc = 500
2
7. The circle will intersect the σ-axis at two of the principal stresses. In this case, we see that one is positive and
the other is negative so they will be σ1 and σ3. The third principal stress is σ2 = 0.
2
8. Calculate the radius of the circle
 σx − σy 
2

 + τxy
 2 
R :=
R = 707.1
τ CW
τ CW
τ1-3
τ 1-2
500
500
y
-500
500
σ3
1000
1500
σ
0
σ1
τ2-3
-500
σ3
500
1000
1500
σ
0
σ2
σ1
2φ
500
x
τ CCW
500
τ CCW
FIGURE 4-1aB
2D and 3D Mohr's Circle Diagrams for Problem 4-1a
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
9. Calculate the principal stresses
4-1a-2
σ1 := σc + R
σ1 = 1207
σ3 := σc − R
σ3 = −207
σ2 := 0
10. Draw the three Mohr's circles to represent the complete 3D stress state.
y
11. Calculate the principal shear stresses
τ12 := 0.5⋅ ( σ1 − σ2)
τ12 = 603.6
τ23 := 0.5⋅ ( σ2 − σ3)
τ23 = 103.6
τ13 := 0.5⋅ ( σ1 − σ3)
τ13 = 707.1
207
1207
22.5°
x
The maximum principal stress is always τ13.
12. Determine the orientation of the principal normal
stress (σ1) with respect to the x-axis. From the 2D
Mohr's circle diagram, we see that the angle 2φ from x
to σ1 is CCW and is given by
 σx − σc 
ϕ := ⋅ acos

2
 R 
1
ϕ = 22.5 deg
FIGURE 4-1aC
Rotated Stress Element for Problem 4-1a
13. Draw the rotated 2D stress element showing the two nonzero principal stresses.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-1h-1
PROBLEM 4-1h
Statement:
A differential stress element has a set of applied stresses on it as indicated in each row of Table
P4-1. For row h, draw the stress element showing the applied stresses, find the principal stresses
and maximum shear stress and draw the Mohr's circle diagram.
Given:
σx := 750
σy := 500
σz := 250
τxy := 500
τyz := 0
τzx := 0
Solution:
See Figures 4-1h and Mathcad file P0401h.
z
250
1. Calculate the coefficients (stress invariants) of equation (4.4c).
3
C2 := σx + σy + σz
C2 = 1.500 × 10
 σx τxy 
 σx τzx 
 σy τyz 

 + 
 + 

 τxy σy 
 τzx σz 
 τyz σz 
C1 :=
750
5
C1 = 4.375 × 10
500
500
x
 σx τxy τzx 


C0 :=  τxy σy τyz 
τ τ σ 
 zx yz z 
7
y
FIGURE 4-1hA
C0 = 3.125 × 10
Stress Element for Problem 4-1h
3
2. Find the roots of the triaxial stress equation:
 −C0 


C1 

v :=
 −C2 


 1 
500
r := polyroots ( v)
2
σ − C2⋅ σ + C 1⋅ σ − C0 = 0
 110 

r = 250 


 1140 
τ CW
3. Extract the principal stresses from
the vector r by inspection.
σ1 := r
σ1 = 1140
3
σ2 := r
σ2 = 250
σ3 := r
σ3 = 110
2
1
τ12 :=
τ23 :=
τ2-3
-500
4. Using equations (4.5), evaluate
the principal shear stresses.
τ13 :=
τ1-3
τ 1-2
500
σ1 − σ3
2
σ1 − σ2
2
σ2 − σ3
2
τ13 = 515
500
1000
1500
σ
0
σ3
σ2
σ1
500
τ12 = 445
τ23 = 70
τ CCW
FIGURE 4-1hB
5. Draw the three-circle Mohr diagram.
The Three Mohr's Circles for Problem 4-1h
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-2-1
PROBLEM 4-2
Statement:
A 400-lb chandelier is to be hung from two 10-ft-long solid steel cables in tension. Choose a
suitable diameter for the cable such that the stress will not exceed 5000 psi. What will be the
deflection of the cables? State all assumptions.
Given:
Weight of chandelier
Length of cable
Allowable stress
W := 400⋅ lbf
L := 10⋅ ft
L = 120 in
σallow := 5000⋅ psi
Number of cables
N := 2
Young's modulus
E := 30⋅ 10 ⋅ psi
6
Assumptions: The cables share the load equally.
Solution:
See Mathcad file P0402.
W
1.
Determine the load on each cable
2.
The stress in each cable will be equal to the load on the cable divided by its cross-sectional area. Using
equation (4.7), and setting the stress equal to the allowable stress, we have
P :=
P = 200 lbf
N
4⋅ P
σallow =
π⋅ d
3.
2
Solve this equation for the unknown cable diameter.
d :=
4⋅ P
d = 0.226 in
π⋅ σallow
4.
Round this up to the next higher decimal equivalent of a common fractional size:
5.
Using equation (4.8), determine the deflection in each cable.
Cross-section area
Cable deflection
A :=
π⋅ d
∆s :=
2
4
P⋅ L
A ⋅E
d := 0.250⋅ in
2
A = 0.049 in
∆s = 0.016 in
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-3-1
PROBLEM 4-3
Statement:
For the bicycle pedal-arm assembly in Figure 4-1 with rider-applied force of 1500 N at the pedal,
determine the maximum principal stress in the pedal arm if its cross-section is 15 mm in dia. The
pedal attaches to the pedal arm with a 12-mm screw thread. What is the stress in the pedal screw?
Given:
Distances (see figure)
Rider-applied force
a  170  mm
Frider  1.5 kN
Pedal arm diameter
d pa  15 mm
Screw thread diameter
d sc  12 mm
b  60 mm
z
Solution:
See Figure 4-3 and Mathcad file P0403.
a
1. From the FBD in Figure 4-3A (and on
the solution for Problem 3-3), we see that
the force from the rider is reacted in the
pedal arm internally by a moment, a
torque, and a vertical shear force. There
are two points at section C (Figure 4-3B)
that we should investigate, one at z = 0.5
d pa (point A), and one at y = 0.5 d pa (point
B).
Frider a  Mc = 0
 M x:
Frider b  Tc = 0
Mc
b
Arm
y
Fc
Pedal
x
2. Refering to the FBD resulting from
taking a section through the arm at C, the
maximum bending moment Mc is found by
summing moments about the y-axis, and
the maximum torque Tc is found by
summing moments about the x-axis.
 M y:
Tc
C
Frider
FIGURE 4-3A
Free Body Diagram for Problem 4-3
z
Section C
A
Maximum bending moment:
Mc  Frider a
Mc  255  N  m
Maximum torque:
Tc  Frider b
Fc  Frider
3.
x
Tc  90 N  m
Vertical shear:
B
Arm
y
FIGURE 4-3B
Fc  1.500  kN
Points A and B at Section C
Determine the stress components at point A where we have the effects of bending and torsion, but where the
transverse shear due to bending is zero because A is at the outer fiber. Looking down the z-axis at a stress
element on the surface at A,
Distance to neutral axis
cpa  0.5 d pa
cpa  7.5 mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4
π d pa
Moment of inertia of
pedal-arm
Ipa 
Bending stress
(x-direction)
σx 
Stress in y-direction
σy  0  MPa
Torsional stress
due to Tc
τxy 
Principal stresses at A,
equation (4.6a)
Mc cpa
Tc cpa
τxy  135.8  MPa
2  Ipa
σx  σy
2
σx  σy
2
CW
2
 σx  σy 
2
 
  τxy
2


2
 σx  σy 
2
 
  τxy
 2 
σ2A  0  MPa
σ3A  23 MPa
Determine the stress components at point B where we have the effects of transverse shear and torsion, but
where the bending stress is zero because B is on the neutral plane. Looking down the y-axis at a stress
element at B,
2
π d pa
Cross-section area
of pedal-arm
Apa 
Torsional stress
due to Tc and shear
stress due to Fc
τzx 
Normal stresses
σx  0  MPa
Principal stresses at B
σ1B  124  MPa
Apa  176.7  mm
4
4 Fc

 τxy
3 Apa
σ1B 
σ3B 
5.
4
σx  769.6  MPa
Ipa
σ1A 
σ1A  793  MPa
3
Ipa  2.485  10  mm
64
σ3A 
4.
4-3-2
τzx  124.5  MPa
CW
σz  0  MPa
σx  σz
2
σx  σz
2
2
 σx  σz 
2
 
  τzx
2


2
 σx  σz 
2
 
  τzx
 2 
σ2B  0  MPa
The maximum principal stress is at point A and is
2
σ3B  124  MPa
σ1A  793  MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6.
4-3-3
Determine the stress in the pedal screw.
Bending moment
Msc  Frider b
Msc  90 N  m
Distance to neutral axis
csc  0.5 d sc
csc  6  mm
Moment of inertia of
pedal screw
Isc 
Bending stress
(y-direction)
σy 
Stress in z-direction
σz  0  MPa
Torsional stress
τxy  0  MPa
4
π d sc
3
Isc  1.018  10  mm
64
Msc csc
Isc
4
σy  530.5  MPa
Since there is no shear stress present at the top of the screw where the bending stress is a maximum, the
maximum principal stress in the pedal screw is
σ1  σy
σ1  530.5  MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-4-1
PROBLEM 4-4
Statement:
The trailer hitch shown in Figure P4-2 and Figure 1-1 (p. 12) has loads applied as defined in
Problem 3-4. The tongue weight of 100 kg acts downward and the pull force of 4905 N acts
horizontally. Using the dimensions of the ball bracket shown in Figure 1-5 (p. 15), determine:
(a) The principal stresses in the shank of the ball where it joins the ball bracket.
(b) The bearing stress in the ball bracket hole.
(c) The tearout stress in the ball bracket.
(d) The normal and shear stresses in the 19-mm diameter attachment holes.
(e) The principal stresses in the ball bracket as a cantilever.
Given:
a  40 mm
b  31 mm
Mtongue  100  kg Fpull  4.905  kN
c  70 mm
d sh  26 mm
d  20 mm
t  19 mm
Assumptions: 1. The nuts are just snug-tight (no pre-load), which is the worst case.
2. All reactions will be concentrated loads rather than distributed loads or pressures.
Solution:
See Figure 4-4 and Mathcad file P0404.
Wtongue  Mtongue g
1. The weight on the tongue is
Wtongue  0.981  kN
2. Solving first for the reactions on the ball by summing the horizontal and vertical forces and the moments
about A.
W tongue
70 = c
1
F pull
1
40 = a
2
A
B
A
19 = t
B
F b1
31 = b
F a1x
C
F a1y
20 = d
F a2y
D
Fa2x
2
Fc2x
F b2
C
D
Fd2
F c2y
FIGURE 4-4A
Dimensions and Free Body Diagram for Problem 4-4
 Fx :
Fpull  Fa1x  Fb1 = 0
(1)
 Fy :
Fa1y  Wtongue = 0
(2)
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
 MA:
4-4-2
Fb1 t  Fpull  a = 0
(3)
Fpull  a
3. Solving equation (3) for Fb1
Fb1 
Fb1  10.326 kN
4. Substituting into (1) and solving for Fa1x
Fa1x  Fpull  Fb1
Fa1x  15.231 kN
5. Solving (2) for Fa1y
Fa1y  Wtongue
Fa1y  0.981  kN
t
6. Now, refering to the FBD of the bracket, we can apply the equations of equilibrium to determine the reactions
at C and D on the bracket.
 Fx :
Fa2x  Fb2  Fc2x  Fd2 = 0
(4)
 Fy :
Fc2y  Fa2y = 0
(5)
 MC:
Fd2 d  Fb2 b  Fa2x ( b  t)  Fa2y c = 0
(6)
7. Note also that the interface forces between part 1 (ball) and part 2 (bracket) have been drawn on their
respective FBDs in opposite senses. Therefore,
Fa2x  Fa1x
Fa2y  Fa1y
8. Solving equation (6) for Fd2
Fd2 
Fb2  Fb1
Fa2x ( b  t)  Fa2y c  Fb2 b
Fd2  25.505 kN
d
9. Substituting into (4) and solving for Fc2x
Fc2x  Fa2x  Fb2  Fd2
Fc2x  30.41  kN
10. Solving (5) for Fa1y
Fc2y  Fa2y
Fc2y  0.981  kN
11. Determine the principal stresses in the shank of the ball where it joins the ball bracket.
The internal bending moment at A on the FBD of the ball is
Distance to neutral axis
Moment of inertia of shank
Bending stress (x-direction)
M  Fpull  a
M  196.2  N  m
csh  0.5 d sh
csh  13 mm
Ish 
σx 
π  d sh
64
M  csh
Ish
Stress in y-direction
σy  0  MPa
Shear stress at A
τxy  0  MPa
4
4
Ish  2.243  10  mm
4
σx  113.7  MPa
Since the shear stress is zero, x is the maximum principal stress, thus
σ1  σx
σ1  114  MPa
σ2  0  MPa
σ3  0  MPa
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-4-3
12. Determine the bearing stress in the ball bracket hole.
Bearing area
Abearing  d sh t
Bearing stress
σbearing 
Abearing  494  mm
Fpull
2
σbearing  9.93 MPa
Abearing
13. Determine the tearout stress in the ball bracket.
Tearout length
Shear area (see Figure 4-4B)
2
Atear = 2  t R  ( 0.5 d )
2
Atear  2  t ( 32 mm)   0.5 d sh
2
Atear  1111 mm
2
2
Stress
R
d
τtear 
Fpull
Atear
FIGURE 4-4B
Tearout Diagram for Problem 4-4
τtear  4.41 MPa
d bolt  19 mm
14. Determine the normal and shear stresses in the attachment bolts if they are 19-mm dia.
Bolt cross-section area (2 bolts)
Abolt  2 
Normal stress (tension)
σbolt 
π d bolt
Fc2x
Abolt
4
2
Abolt  567.1  mm
2
σbolt  53.6 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,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
Shear stress
τbolt 
4-4-4
W tongue
Fc2y
τbolt  1.7 MPa
Abolt
1
F pull
15. Determine the principal stresses in the ball bracket as a
cantilever (see Figure 4-4C).
a
Bending moment
2
M  Fpull  a  Wtongue c
Width of bracket
Moment of inertia
Total tensile stress
M  264.8  N  m
M
w  64 mm
I 
w t
R
3
I  36581  mm
12
σ 
c
M t
2 I

Fpull
w t
4
FIGURE 4-4C
Cantilever FBD for Problem 4-4
σ  72.8 MPa
Since there are no shear stress at the top and bottom of the bracket where the bending stresses are maximum,
they are also the principal stresses, thus
σ1  σ
τmax 
σ1  72.8 MPa
σ
2
σ2  0  MPa
σ3  0  MPa
τmax  36.4 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,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-5-1
PROBLEM 4-5
Statement:
Repeat Problem 4-4 for the loading conditions of Problem 3-5, i.e., determine the stresses due to a
horizontal force that will result on the ball from accelerating a 2000-kg trailer to 60 m/sec in 20 se
Assume a constant acceleration. From Problem 3-5, the pull force is 6000 N. Determine:
(a) The principal stresses in the shank of the ball where it joins the ball bracket.
(b) The bearing stress in the ball bracket hole.
(c) The tearout stress in the ball bracket.
(d) The normal and shear stresses in the 19-mm diameter attachment holes.
(e) The principal stresses in the ball bracket as a cantilever.
Given:
a  40 mm
b  31 mm
Mtongue  100  kg Fpull  6  kN
c  70 mm
d sh  26 mm
d  20 mm
t  19 mm
Assumptions: 1. The nuts are just snug-tight (no pre-load), which is the worst case.
2. All reactions will be concentrated loads rather than distributed loads or pressures.
Solution:
See Figure 4-5 and Mathcad file P0405.
1. The weight on the tongue is
Wtongue  Mtongue g
Wtongue  0.981  kN
2. Solving first for the reactions on the ball by summing the horizontal and vertical forces and the moments about
A.
W tongue
70 = c
1
F pull
1
40 = a
2
A
B
A
19 = t
B
F b1
31 = b
F a1x
C
F a1y
20 = d
F a2y
D
Fa2x
2
Fc2x
F b2
C
D
Fd2
F c2y
FIGURE 4-5A
Dimensions and Free Body Diagram for Problem 4-5
 Fx :
Fpull  Fa1x  Fb1 = 0
(1)
 Fy :
Fa1y  Wtongue = 0
(2)
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
 MA:
4-5-2
Fb1 t  Fpull  a = 0
(3)
Fpull  a
3. Solving equation (3) for Fb1
Fb1 
Fb1  12.632 kN
4. Substituting into (1) and solving for Fa1x
Fa1x  Fpull  Fb1
Fa1x  18.632 kN
5. Solving (2) for Fa1y
Fa1y  Wtongue
Fa1y  0.981  kN
t
6. Now, refering to the FBD of the bracket, we can apply the equations of equilibrium to determine the reactions
at C and D on the bracket.
 Fx :
Fa2x  Fb2  Fc2x  Fd2 = 0
(4)
 Fy :
Fc2y  Fa2y = 0
(5)
 MC:
Fd2 d  Fb2 b  Fa2x ( b  t)  Fa2y c = 0
(6)
7. Note also that the interface forces between part 1 (ball) and part 2 (bracket) have been drawn on their respective
FBDs in opposite senses. Therefore,
Fa2x  Fa1x
Fa2y  Fa1y
8. Solving equation (6) for Fd2
Fd2 
Fb2  Fb1
Fa2x ( b  t)  Fa2y c  Fb2 b
Fd2  30.432 kN
d
9. Substituting into (4) and solving for Fc2x
Fc2x  Fa2x  Fb2  Fd2
Fc2x  36.432 kN
10. Solving (5) for Fa1y
Fc2y  Fa2y
Fc2y  0.981  kN
11. Determine the principal stresses in the shank of the ball where it joins the ball bracket.
The internal bending moment at A on the FBD of the ball is
Distance to neutral axis
Moment of inertia of shank
Bending stress (x-direction)
M  Fpull  a
M  240  N  m
csh  0.5 d sh
csh  13 mm
Ish 
σx 
π  d sh
4
64
M  csh
Ish
Stress in y-direction
σy  0  MPa
Shear stress at A
τxy  0  MPa
4
Ish  2.243  10  mm
4
σx  139.1  MPa
Since the shear stress is zero, x is the maximum principal stress, thus
σ1  σx
σ1  139  MPa
12. Determine the bearing stress in the ball bracket hole.
σ2  0  MPa
σ3  0  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,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
Bearing area
Abearing  d sh t
Bearing stress
σbearing 
4-5-3
Abearing  494  mm
Fpull
2
σbearing  12.15  MPa
Abearing
13. Determine the tearout stress in the ball bracket.
Tearout length
Shear area (see Figure 4-4B)
2
Atear = 2  t R  ( 0.5 d )
2
Atear  2  t ( 32 mm)   0.5 d sh
2
Atear  1111 mm
2
2
Stress
R
d
τtear 
Fpull
Atear
FIGURE 4-5B
τtear  5.4 MPa
Tearout Diagram for Problem 4-5
d bolt  19 mm
14. Determine the normal and shear stresses in the attachment bolts if they are 19-mm dia.
Bolt cross-section area (2 bolts)
Abolt  2 
σbolt 
Normal stress (tension)
π d bolt
2
Abolt  567.1  mm
4
Fc2x
σbolt  64.2 MPa
Abolt
Shear stress
τbolt 
2
W tongue
Fc2y
τbolt  1.7 MPa
Abolt
1
F pull
15. Determine the principal stresses in the ball bracket as a
cantilever (see Figure 4-4C).
a
Bending moment
2
M  Fpull  a  Wtongue c
Width of bracket
Moment of inertia
M  308.6  N  m
M
w  64 mm
I 
w t
c
R
3
I  36581  mm
12
4
FIGURE 4-5C
Cantilever FBD for Problem 4-5
Total tensile stress
σ 
M t
2 I

Fpull
w t
σ  85.1 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,
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Pearson Education, Inc., Upper Saddle River, NJ 07458.
MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-5-4
Since there are no shear stress at the top and bottom of the bracket where the bending stresses are maximum,
they are also the principal stresses, thus
σ1  σ
τmax 
σ1  85.1 MPa
σ
2
σ2  0  MPa
σ3  0  MPa
τmax  42.5 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,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-6-1
PROBLEM 4-6
Statement:
Repeat Problem 4-4 for the loading conditions of Problem 3-6, i.e., determine the stresses due to a
horizontal force that will results from an impact between the ball and the tongue of the 2000-kg
trailer if the hitch deflects 2.8 mm dynamically on impact. The tractor weighs 1000 kg and the
velocity at impact is 0.3 m/sec. Determine:
(a) The principal stresses in the shank of the ball where it joins the ball bracket.
(b) The bearing stress in the ball bracket hole.
(c) The tearout stress in the ball bracket.
(d) The normal and shear stresses in the 19-mm diameter attachment holes.
(e) The principal stresses in the ball bracket as a cantilever.
Given:
a  40 mm
b  31 mm
Mtongue  100  kg Fpull  55.1 kN
c  70 mm
d sh  26 mm
d  20 mm
t  19 mm
Assumptions: 1. The nuts are just snug-tight (no pre-load), which is the worst case.
2. All reactions will be concentrated loads rather than distributed loads or pressures.
Solution:
See Figure 4-6 and Mathcad file P0406.
1. The weight on the tongue is
Wtongue  Mtongue g
Wtongue  0.981  kN
2. Solving first for the reactions on the ball by summing the horizontal and vertical forces and the moments
about A.
W tongue
70 = c
1
F pull
1
40 = a
2
A
B
A
19 = t
B
F b1
31 = b
F a1x
C
F a1y
20 = d
F a2y
D
Fa2x
2
Fc2x
F b2
C
D
Fd2
F c2y
FIGURE 4-6A
Dimensions and Free Body Diagram for Problem 4-6
 Fx :
Fpull  Fa1x  Fb1 = 0
(1)
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-6-2
 Fy :
Fa1y  Wtongue = 0
(2)
 MA:
Fb1 t  Fpull  a = 0
(3)
Fpull  a
3. Solving equation (3) for Fb1
Fb1 
Fb1  116  kN
4. Substituting into (1) and solving for Fa1x
Fa1x  Fpull  Fb1
Fa1x  171.1  kN
5. Solving (2) for Fa1y
Fa1y  Wtongue
Fa1y  0.981  kN
t
6. Now, refering to the FBD of the bracket, we can apply the equations of equilibrium to determine the reactions
at C and D on the bracket.
 Fx :
Fa2x  Fb2  Fc2x  Fd2 = 0
(4)
 Fy :
Fc2y  Fa2y = 0
(5)
 MC:
Fd2 d  Fb2 b  Fa2x ( b  t)  Fa2y c = 0
(6)
7. Note also that the interface forces between part 1 (ball) and part 2 (bracket) have been drawn on their respective
FBDs in opposite senses. Therefore,
Fa2x  Fa1x
Fa2y  Fa1y
8. Solving equation (6) for Fd2
Fd2 
Fb2  Fb1
Fa2x ( b  t)  Fa2y c  Fb2 b
Fd2  251.382  kN
d
9. Substituting into (4) and solving for Fc2x
Fc2x  Fa2x  Fb2  Fd2
Fc2x  306.482  kN
10. Solving (5) for Fa1y
Fc2y  Fa2y
Fc2y  0.981  kN
11. Determine the principal stresses in the shank of the ball where it joins the ball bracket.
The internal bending moment at A on the FBD of the ball is
Distance to neutral axis
Moment of inertia of shank
Bending stress (x-direction)
3
M  Fpull  a
M  2.204  10  N  m
csh  0.5 d sh
csh  13 mm
Ish 
σx 
π  d sh
64
M  csh
Ish
Stress in y-direction
σy  0  MPa
Shear stress at A
τxy  0  MPa
4
4
Ish  2.243  10  mm
4
σx  1277 MPa
Since the shear stress is zero, x is the maximum principal stress, thus
σ1  σx
σ1  1277 MPa
σ2  0  MPa
σ3  0  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,
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Pearson Education, Inc., Upper Saddle River, NJ 07458.
MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-6-3
12. Determine the bearing stress in the ball bracket hole.
Bearing area
Abearing  d sh t
Bearing stress
σbearing 
Abearing  494  mm
Fpull
2
σbearing  111.54 MPa
Abearing
13. Determine the tearout stress in the ball bracket.
Tearout length
Shear area (see Figure 4-4B)
2
Atear = 2  t R  ( 0.5 d )
2
Atear  2  t ( 32 mm)   0.5 d sh
2
Atear  1111 mm
2
2
Stress
R
d
τtear 
Fpull
Atear
FIGURE 4-6B
Tearout Diagram for Problem 4-6
τtear  49.59  MPa
d bolt  19 mm
14. Determine the normal and shear stresses in the attachment bolts if they are 19-mm dia.
Bolt cross-section area (2 bolts)
Abolt  2 
σbolt 
Normal stress (tension)
π d bolt
2
Abolt  567.1  mm
4
Fc2x
σbolt  540  MPa
Abolt
Shear stress
τbolt 
2
W tongue
Fc2y
τbolt  1.7 MPa
Abolt
1
F pull
15. Determine the principal stresses in the ball bracket as a
cantilever (see Figure 4-4C).
a
Bending moment
Width of bracket
Moment of inertia
M  2.3  10  N  m
M
w  64 mm
I 
2
3
M  Fpull  a  Wtongue c
w t
c
R
3
I  36581  mm
12
4
FIGURE 4-6C
Cantilever FBD for Problem 4-6
Total tensile stress
σ 
M t
2 I

Fpull
w t
σ  635.5  MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-6-4
Since there are no shear stress at the top and bottom of the bracket where the bending stresses are maximum,
they are also the principal stresses, thus
σ1  σ
τmax 
σ1  636  MPa
σ
2
σ2  0  MPa
σ3  0  MPa
τmax  318  MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-7-1
PROBLEM 4-7
Statement:
Design the wrist pin of Problem 3-7 for a maximum allowable principal stress of 20 ksi if the pin is
hollow and loaded in double shear.
Given:
Force on wrist pin
Fwristpin  12.258 kN
Allowable stress
σallow  20 ksi
od  0.375  in
Assumptions: Choose a suitable outside diameter, say
Solution:
Fwristpin  2756 lbf
See Figure 4-12 in the text and Mathcad file P0407.
1. The force at each shear plane is
F 
Fwristpin
F  1378 lbf
2
2. With only the direct shear acting on the plane, the Mohr diagram will be a circle with center at the origin and
radius equal to the shear stress. Thus, the principal normal stress is numerically equal to the shear stress, which
in this case is also the principal shear stress, so we have  = 1 = allow.
3. The shear stress at each shear plane is
4. Solving for the inside diameter,
τ=
id 
F
A
2
=
od 
4 F
2
2
π  od  id 
4 F
π σallow
= σallow
id  0.230  in
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-8-1
PROBLEM 4-8
Statement:
A paper mill processes rolls of paper having a density of 984 kg/m3. The paper roll is 1.50 m
outside dia (OD) by 0.22 m inside dia (ID) by 3.23 m long and is on a simply supported, hollow,
steel shaft. Find the shaft ID needed to obtain a maximum deflection at the center of 3 mm if the
shaft OD is 22 cm.
Given:
Paper density
ρ  984 
kg
3
m
Roll dimensions
Outside diameter
Inside diameter
Lemgth
od  220  mm
E  207  GPa
δ  3  mm
Shaft outside dia
Young's modulus
Allowable deflection
OD  1.50 m
ID  0.22 m
L  3.23 m
Assumptions: The shaft (beam) supporting the paper roll is simply-supported at the ends and is the same
length as the paper roll. The paper acts as a distributed load over the length of the shaft.
Solution:
1.
See Mathcad file P0408.
The weight of the paper roll is equal to its volume times the paper density times g.
Wroll 

4
π
2

2
 OD  ID  L ρ  g
Wroll  53.89  kN
Wroll
The intensity of the distributed load is w 
3.
Using Figure B-2(b) in Appendix B with a = 0, the maximum deflection is at the midspan and is
y =
w x
24 E I

2
3
w  16.686
N
2.
L

3
 2  L x  x  L
mm
4
For x = L/2, this reduces to
y =
5  w L
384  E I
4
Letting  = -y and solving for I, we have
4.
I 
5  w L
7
I  3.808  10  mm
384  E δ
The area moment of inertia for a hollow circular cross-section is I =
π
64

4
4

4
 od  id
1
Solving this for the id yields
id   od 
4

Round this down (for slightly less deflection) to
64 I 
4

π 
id  198.954  mm
id  198  mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-9-1
PROBLEM 4-9
Statement:
A ViseGrip plier-wrench is drawn to scale in Figure P4-3, and for which the forces were analyzed
in Problem 3-9, find the stresses in each pin for an assumed clamping force of P = 4000 N in the
position shown. The pins are 8-mm dia and are all in double shear.
Given:
Pin forces as calculated in Problem 3-9:
Member 1
F21  7.5 kN
F41  5.1 kN
Member 2
F12  7.5 kN
F32  5.1 kN
Member 3
F23  5.1 kN
F43  5.1 kN
Member 4
F14  5.1 kN
F34  5.1 kN
d  8  mm
Pin diameter
Assumptions: Links 3 and 4 are in a toggle position, i.e., the pin that joins links 3 and 4 is in line with the pins
that join 1 with 4 and 2 with 3.
Solution:
1.
See Figure 4-9 and Mathcad file P0409.
The FBDs of the assembly and each individual link are shown in Figure 4-9. The dimensions, as scaled from
Figure P4-3 in the text, are shown on the link FBDs.
4
F
P
1
2
3
P
F
55.0 = b
50.0 = a
39.5 = c
F
F14
22.0 = d
129.2°
1

4
F34
F41
F21
P

28.0 = e


F43

F12
3
F23
P
F32
2.8 = g
21.2 = h
2
F
26.9 = f
FIGURE 4-9
Free Body Diagrams for Problem 4-9
2.
The cross-sectional area for all pins is the same and is
A 
π d
4
2
A  50.265 mm
2
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3.
4-9-2
The pin that joins members 1 and 2 is the most highly stressed while the stress on each of the remaining pins
is the same. Since the pins are in double shear, we will divide the pin load by 2 in each case.
Pin joining 1 and 2
τ12 
All other pins
τ14 
F12
2 A
F14
2 A
τ12  74.6 MPa
τ14  50.7 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-10-1
PROBLEM 4-10
Statement:
Given:
The over-hung diving board of problem 3-10 is shown in Figure P4-4a. Assume cross-section
dimensions of 305 mm x 32 mm. The material has E = 10.3 GPa. Find the largest principal stress
at any location in the board when a 100-kg person is standing at the free end.
W  100  kgf
Weight of person
2000 = L
Board dimensions
R1
Distance to support
a  0.7 m
Length of board
L  2  m
Cross-section
w  305  mm
P
R2
t  32 mm
700 = a
FIGURE 4-10
Assumptions: The weight of the beam is negligible
compared to the applied load and so can
be ignored.
Solution:
1.
See Figure 4-10 and Mathcad file P0410.
From the FBD of the diving board and Figure B-3 (Appendix B), the reactions at the supports are
R1  W   1 

R2  W  
L

R1  1821 N
a
L

a
2.
R2  2802 N
Also from Figure D-3, the maximum bending moment occurs at the right-hand support where, in the FBD
above, x = a.
Mmax  R1 a
3.
Free Body Diagram for Problem 4-10
Mmax  1275 N  m
The maximum bending stress will occur on the top and bottom surfaces of the board at the section where the
maximum bending moment occurs which, in this case, is at x = a. The only stress present on the top or
bottom surface of the board is the bending stress x. Therefore, on the top surface where the stress is
tensile, x. is the principal stress 1 . Thus,
Distance to extreme fiber
Moment of inertia
Bending stress
Maximum principal stress
c 
I 
σx 
t
c  16 mm
2
w t
3
12
Mmax c
σ1  σx
5
I  8.329  10  mm
I
4
σx  24.492 MPa
σ1  24.5 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-11-1
PROBLEM 4-11
Statement:
Given:
Repeat Problem 4-10 using the loading conditions of Problem 3-11. Assume the board weighs 29
kg and deflects 13.1 cm statically when the person stands on it. Find the largest principal stress
at any location in the board when the 100-kg person in Problem 4-10 jumps up 25 cm and lands
back on the board. Find the maximum deflection.
Beam length
L  2000 mm
Distance to support
a  700  mm
Mass of person
mpers  100  kg
Mass of board
mboard  29 kg
Static deflection
δst  131  mm
Height of jump
h  250  mm
Cross-section
w  305  mm
2000 = L
R1
Fi
R2
700 = a
FIGURE 4-11
t  32 mm
Free Body Diagram for Problem 4-11
Assumptions: The apparent Young's modulus for fiberglas is
4
E  1.03 10  MPa
Solution:
See Figure 4-11 and Mathcad file P0411.
1. From Problem 3-11, the dynamic load resulting from the impact of the person with the board isFi  3.056  kN
2. From the FBD of the diving board and Figure B-3(a) (Appendix B), the reactions at the supports are
R1  Fi  1 

R2  Fi 
L

R1  5.675  kN
a
L

a
R2  8.731  kN
3. Also from Figure D-3(a), the maximum bending moment occurs at the right-hand support where, in the
FBD above, x = a.
Mmax  R1 a
Mmax  3.973  kN  m
4. The maximum bending stress will occur on the top and bottom surfaces of the board at the section where the
maximum bending moment occurs which, in this case, is at x = a. The only stress present on the top or bottom
surface of the board is the bending stress x. Therefore, on the top surface where the stress is tensile, x is the
principal stress 1 . Thus,
c 
Distance to extreme fiber
I 
Moment of inertia
σx 
Bending stress
t
c  16 mm
2
w t
3
12
Mmax c
I
σ1  σx
Maximum principal stress
5
I  8.329  10  mm
4
σx  76.322 MPa
σ1  76.3 MPa
5. Calculate the maximum deflection from the equation given in Figure D-3(a) at x = L. Let b in the figure be a in
our problem and let a in the figure be equal to L, then
ymax 
Fi
6  a  E I
 ( a  L)  L  L ( L  a )  a  ( L  a )  L
3
3
2
ymax  401.4  mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-12-1
PROBLEM 4-12
Statement:
Repeat Problem 4-10 using the cantilevered diving board design in Figure P4-4b.
Given:
Beam length
L  1300 mm
Weight at free end
P  100  kgf
Cross-section
w  305  mm
2000
1300 = L
P
t  32 mm
Assumptions:
The apparent Young's modulus for
fiberglas is
M1
R1
4
E  1.03 10  MPa
Solution:
700
FIGURE 4-12
See Figure 4-12 and Mathcad file P0412.
Free Body Diagram for Problem 4-12
1.
2.
From the FBD of the diving board and Figure B-1(a) (Appendix B), the reactions at the supports are
R1  P
R1  981  N
M1  P L
M1  1275 N  m
Also from Figure D-1, the maximum bending moment occurs at the support where, in the FBD above, x = 0.
Mmax  M1
Mmax  1275 N  m
3. The maximum bending stress will occur on the top and bottom surfaces of the board at the section where the
maximum bending moment occurs which, in this case, is at x = 0. The only stress present on the top or bottom
surface of the board is the bending stress x. Therefore, on the top surface where the stress is tensile, x is the
principal stress 1 . Thus,
Distance to extreme fiber
Moment of inertia
Bending stress
Maximum principal stress
c 
I 
σx 
t
c  16 mm
2
w t
3
12
Mmax c
σ1  σx
5
I  8.329  10  mm
I
4
σx  24.492 MPa
σ1  24.5 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-13-1
PROBLEM 4-13
Statement:
Given:
Repeat Problem 4-11 using the diving board design shown in Figure P4-4b. Assume the board
weighs 19 kg and deflects 8.5 cm statically when the person stands on it.
Unsupported length
L  1300 mm
Mass of board
mboard  19 kg
Static board deflection
δstat  85 mm
Mass of person
mperson  100  kg
Height of jump
h  250  mm
Cross-section
w  305  mm
2000
1300 = L
Fi
M1
t  32 mm
R1
700
Assumptions: The apparent Young's modulus for fiberglas is
FIGURE 4-13
4
E  1.03 10  MPa
Solution:
Free Body Diagram for Problem 4-13
See Figure 4-13 and Mathcad file P0413.
1.
From Problem 3-13, the dynamic load resulting from the impact of the person with the board isFi  3.487  kN
2.
From the FBD of the diving board and Figure B-1(a) (Appendix B), the reactions at the supports are
3.
R1  Fi
R1  3487 N
M1  Fi L
M1  4533 N  m
Also from Figure D-1, the maximum bending moment occurs at the support where, in the FBD above, x = 0.
Mmax  M1
Mmax  4533 N  m
4. The maximum bending stress will occur on the top and bottom surfaces of the board at the section where the
maximum bending moment occurs which, in this case, is at x = 0. The only stress present on the top or bottom
surface of the board is the bending stress x. Therefore, on the top surface where the stress is tensile, x is the
principal stress 1 . Thus,
Distance to extreme fiber
Moment of inertia
Bending stress
Maximum principal stress
5.
c 
I 
σx 
t
c  16 mm
2
w t
3
5
I  8.329  10  mm
12
Mmax c
4
σx  87.086 MPa
I
σ1  σx
σ1  87.1 MPa
Calculate the maximum deflection from the equation given in Figure D-3(a) at x = L. Let b in the figure be a in
our problem and let a in the figure be equal to L, then
3
ymax 
Fi L
3  E I
ymax  297.7  mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-14-1
PROBLEM 4-14
Statement:
Figure P4-5 shows a child's toy called a pogo stick. The child stands on the pads, applying
half her weight on each side. She jumps off the ground, holding the pads up against her
feet, and bounces along with the spring cushioning the impact and storing energy to help
each rebound. Design the aluminum cantilever beam sections on which she stands to
survive jumping 2 in off the ground. Assume an allowable stress of 20 ksi. Define and size
the beam shape.
Given:
Allowable stress
σallow  20 ksi
Young's modulus
E  10.3 10  psi
6
Assumptions: The beam will have a rectangular
cross-section with the load applied at a
distance of 5 in from the central support.
L  5  in
Solution:
See Figure 4-14 and Mathcad file P0414.
1. From Problem 3-14, the total dynamic force on both
foot supports is
Fi  224  lbf
Fi /2
Therefore, the load on each support is
P 
Fi
Fi /2
P  112  lbf
2
2. To give adequate support to the childs foot, let the
width of the support beam be
w  1.5 in
3. From Figure B-1(a) in Appendix B, the maximum
bending moment at x = 0 is
M  P L
4.
We can now calculate the minimum required section modulus, Z = I/c.
Bending stress
Solving for Z,
5.
P
M  560  in lbf
σ=
Z 
For a rectangular cross-section, I =
Solving for t,
M
M
t 
Z  458.8  mm
σallow
3
12
Free Body Diagram for Problem 4-14
= σallow
Z
w t
FIGURE 4-14
and c =
6 Z
w
t
2
so Z =
w t
3
2
6
t  0.335  in
Round this up to the next higher decimal equivalent of a common fraction,
t  0.375  in
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-15-1
PROBLEM 4-15
Statement:
Design a shear pin for the propeller shaft of an outboard motor if the shaft through which the
pin is placed is 25-mm diameter, the propeller is 20-cm diameter, and the pin must fail when a
force > 400 N is applied to the propeller tip. Assume an ultimate shear strength for the pin
material of 100 MPa.
Given:
Propeller shaft dia
Propeller dia
Max propeller tip force
d  25 mm
D  200  mm
Fmax  400  N
Ultimate shear strength
S us  100  MPa
Fpin
T
Propeller Hub
Shear Pin
Assumptions: A shear pin is in direct, double shear.
Solution:
Fpin
See Figure 4-15 and Mathcad file P0415.
d
FIGURE 4-15
1. Calculate the torque on the propeller shaft that will
result from a tip force on the propeller of Fmax.
T  Fmax
2.
2
Free Body Diagram for Problem 4-15
T  40000  N  mm
This will be reacted by the shear pin's couple on the shaft. Determine the magnitude of the direct shear force.
Fpin 
3.
D
Propeller Shaft
T
d
Fpin  1600 N
Determine the maximum pin diameter that will shear at this force.
Direct shear stress
τ=
Fpin
A
=
4  Fpin
π d pin
4  Fpin
Solving for the pin diameter
d pin 
Round this to
d pin  4.5 mm
π S us
2
= S us
d pin  4.514  mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-16-1
PROBLEM 4-16
Statement:
Given:
A track to guide bowling balls is designed with two round rods as shown in Figure P4-6. The
rods are not parallel to one another but have a small angle between them. The balls roll on the
rods until they fall between them and drop onto another track. The angle between the rods is
varied to cause the ball to drop at different locations. Find the maximum stress and deflection
in the rods assuming that they are
(a) simply supported at each end, and
(b) fixed at each end.
Rod length
L  30 in
Rod diameter
d  1.00 in
Distance to load
a  23.15  in
Young's modulus
E  30 10  psi
Fball
a
6
R1
Assumptions: The analysis of Problem 3-16 yielded
the following for a simply supported
beam:
R2
L
FIGURE 4-16A
Free Body Diagram for Problem 4-16(a), taken
on a plane through the rod axis and ball center
Max ball load
Fball  13.89  lbf
Max moment
Mmax  73.4 in lbf
Reactions
R1  3.17 lbf
R2  10.72  lbf
Solution:
See Figure 4-16 and Mathcad file P0416.
1. The maximum bending stress will occur at the outer fibers of the rod at the section where the maximum
bending moment occurs which, in this case, is at x = a. The only stress present on the top or bottom surface of
the rod is the bending stress x. Therefore, on the bottom surface where the stress is tensile, x is the principal
stress 1 . Thus, for a simply supported rod,
c 
Distance to extreme fiber
I 
Moment of inertia
σx 
Bending stress
c  0.5 in
2
π d
4
4
I  0.0491 in
64
Mmax c
σx  748  psi
I
σ1  σx
Maximum principal stress
2.
d
σ1  748  psi
Calculate the maximum deflection for the simply supported case from the equation given in Figure D-2(a),
ymax 
Fball
6  E I

  2 a 

3
a
4
L
 L a
2


3. For the case where the rod is built in at each end,
the beam is statically indeterminate. As seen in
Figure 4-16B, there are four unknown reactions and
only two equilibrium equations can be written using
statics. We will find the reactions using Example 4-7
as a model.
ymax  0.0013 in
Fball
a
M1
R1
L
R 2 M2
FIGURE 4-16B
Free Body Diagram for Problem 4-16(b), taken on a
plane through the rod axis and ball center
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-16-2
4. Write an equation for the load function in terms of equations 3.17 and integrate the resulting function four
times using equations 3.18 to obtain the shear and moment functions. Note use of the unit doublet function to
represent the moment at the wall. For the beam in Figure 4-16B,
q(x) = -M1<x - 0>-2 + R1<x - 0>-1 - F<x - a>-1 + R2<x - L>-1 + M2<x - L>-2
V(x) = -M1<x - 0>-1 + R1<x - 0>0 - F<x - a>0
+ R2<x
- L>0 + M2<x - L>-1 + C1
M(x) = -M1<x - 0>0 + R1<x - 0>1 - F<x - a>1 + R2<x - L>1 + M2<x - L>0 + C1x+ C2
(x) = ( -M1<x - 0>1 + R1<x - 0>2/2 - F<x - a>2/2
+ R2<x
- L>2/2 + M2<x - L>1 + C1x2/2 + C2x + C3) / EI
y(x) = ( -M1<x - 0>2/2 + R1<x - 0>3/6 - F<x - a>3/6 + R2<x - L>3/6 + M2<x - L>2 /2+ C1x3/6 + C2x2/2 + C3x + C4) / EI
5. Because the reactions have been included in the loading function, the shear and moment diagrams both
close to zero at each end of the beam, making C1 = C2 = 0. This leaves six unknowns; the four reactions and the
constants of integration, C3 and C4. There are four boundary conditions that we can use and two equilibrium
equations. The boundary conditions are: at x = 0,  = 0 and y = 0; and at x = L,  = 0 and y = 0. Applying the
boundary conditions at x = 0 results in C3 = C4 = 0. Applying the BCs at x = L results in the following two
equations, which are solved for R1 and M1.
At x = L,
θ=0
0=
y =0
0=
R1
2
R1
6
2
 L  M 1 L 
3
L 
M1
2
2
F
2
L 
 ( L  a)
F
6
2
 ( L  a)
3
Solving these two equations simultaneously for R1 and M1,
M1 
Fball
R1  2 
6.
L
M1
L

( L  a)
 ( L  a ) 
2

 Fball 
L
( L  a)
3


M1  16.765 in lbf
2
2
R1  1.842  lbf
L
The remaing two reactions can be found by using the equations of equilibrium.
 Fy = 0:
R1  Fball  R2 = 0
 M = 0:
M1  Fball  a  R2 L  M2 = 0
Solving these two equations simultaneously for R2 and M2,
7.
R2  Fball  R1
R2  12.048 lbf
M2  M1  Fball  a  R2 L
M2  56.657 in lbf
Define the range for x, x  0  in 0.005  L  L
8. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
S ( x z)  if ( x  z 1 0 )
9. Write the shear, moment, slope, and deflection equations in Mathcad form, using the function S as a multiplying
factor to get the effect of the singularity functions.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-16-3
V ( x)  R1 S ( x 0  mm)  Fball  S ( x a )  R2 S ( x L)
M ( x)  M1 S ( x 0  mm)  R1 S ( x 0  mm)  ( x  0  mm)  Fball  S ( x a )  ( x  a ) 
 M2 S ( x L)  R2 S ( x L)  ( x  L)
θ ( x) 
y ( x) 

1
E I
1
E I
 M1 S ( x 0  mm)  x 
R1
2
2
 S ( x 0  mm)  ( x  0  mm) 
Fball
2

R
 M2 S( x L)  ( x  L)  2  S( x L)  ( x  L) 2
2

 M1

2
 S ( x 0  mm)  x 
2
M2
R1
6
3
 S ( x 0  mm)  ( x  0  mm) 



Fball

R
  S ( x L)  ( x  L) 2  2  S ( x L)  ( x  L) 3
6
 2

2
 S ( x a )  ( x  a ) 
6
3

 S ( x a )  ( x  a ) 



10. Plot the shear, moment, slope, and deflection diagrams.
(a) Shear Diagram
(b) Moment Diagram
5
40
20
Moment, M - lb in
Shear, V - lb
0
5
 10
 15
0
 20
 40
0
10
20
 60
30
0
Distance along beam, x - in
30
(d) Deflection Diagram
0
Deflection - thousandths of in
0.1
Slope - Thousands of Rad
20
Distance along beam, x - in
(c) Slope Diagram
0
 0.1
10
0
10
20
30
 0.2
 0.4
 0.6
 0.8
Distance along beam, x - in
0
10
20
30
Distance along beam, x - in
FIGURE 4-16C
Shear, Moment, Slope, and Deflection Diagrams for Problem 4-16(b)
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-16-4
11 The maximum moment occurs at x = L and is
Mmax  M2
Mmax  56.7 in lbf
12 Calculate the maximum bending and principal stresses.
σx 
Bending stress
Mmax c
I
σ1  σx
Maximum principal stress
σx  577  psi
σ1  577  psi
13. To find the maximum deflection, first determine at what point on the beam the slope is zero. Let this be at x = e.
From the slope diagram, we see that e < a. Using the slope equation and setting it equal to zero, we have
For  = 0
0 = M1 e 
Solving for e
e 
Maximum deflection
2  M1
R1
ymax  y ( e)
R1 2
e
2
e  18.204 in
ymax  0.00063  in
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-17-1
PROBLEM 4-17
Statement:
A pair of ice tongs is shown in Figure P4-7. The ice weighs 50 lb and is 10 in wide across the
tongs. The distance between the handles is 4 in, and the mean radius r of the tong is 6 in. The
rectangular cross-sectional dimensions are 0.75 x 0.312 in. Find the stress in the tongs.
Given:
Assumptions:
beam.
Mean radius of tong
rc  6.00 in
Tong width
w  0.312  in
Tong depth
h  0.75 in
F
C
FC
O
The tong can be analyzed as a curved
11.0 = ax
See Problem 3-17, Figure 4-17,
Solution:
and Mathcad file P0417.
3.5 = cy
FO
2.0 = cx
A
12.0 = by
5.0 = bx
1. The maximum bending moment and axial force in the
tong were found in Problem 3-17 at point A. They are
FB
B
Maximum moment
MA  237.5  in lbf
Axial force at D
FAn  25 lbf
W/2
FIGURE 4-17
Free Body Diagram for Problem 4-17
2.
3.
Calculate the section area, inside radius and outside radus.
A  h  w
A  0.234  in
Inside and outside radii
of section
ri  rc  0.5 h
ri  5.625  in
ro  rc  0.5 h
ro  6.375  in
Use the equation in the footnote on page 195 of the text to calculate the radius of the neutral axis.
Radius of neutral axis
4.
rn 
ro  ri
 ro 
ln 
 ri 
rn  5.992  in
Calculate the eccentricty and the distances from the neutral axis to the extreme fibers.
Eccentricity
e  rc  rn
e  0.007821 in
Distances from neutral
axis to extreme fibers
ci  rn  ri
ci  0.3672 in
co  ro  rn
co  0.3828 in
Stresses at inner and
outer radii
σi 
MA c i
FAn
 
e A ri
A
 MA co  FAn
 
A
 e A ro 
σo  
5.
2
Area of section
σi  8.58 ksi
σo  7.69 ksi
The shear stress is zero at the outer fibers. Therefore, these are the principal stresses. At the inner surface
σ1  σi
σ1  8.58 ksi
σ2  0  ksi
σ3  0  ksi
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-18-1
PROBLEM 4-18
Statement:
A set of steel reinforcing rods is to be stretched axially in tension to create a tensile stress of 30
ksi prior to being cast in concrete to form a beam. Determine how much force will be required to
stretch them the required amount and how much deflection is required. There are 10 rods; each
is 0.75-in diameter and 30 ft long.
Given:
Desired stress
σ  30 ksi
Rod diameter
d  0.75 in
Number of rods
Rod length
Nrods  10
L  30 ft
Young's modulus
E  30 10  psi
6
Assumptions: The rods share the load equally.
Solution:
See Mathcad file P0418.
π d
2
2
1.
Calculate the cross-sectional area of one rod. A 
2.
Determine the force required to achieve the desired stress level in one rod.
σ=
3.
F
A
4
F  13.254 kip
Determine the total force required to achieve the desired stress level in all rods.
Ftotal  Nrods F
4.
F  σ  A
A  0.442  in
Ftotal  132.5  kip
Determine the amount the rods will deflect under the applied load.
δ 
F L
A E
δ  0.360  in
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-19-1
PROBLEM 4-19
Statement:
The clamping fixture used to pull the rods in Problem 4-18 is conected to the hydraulic ram by a
clevis like that shown in Figure P4-8. Determine the size of the clevis pin needed to withstand
the applied force. Assume an allowable shear stress of 20 000 psi and an allowable normal
stress of 40 000 psi. Determine the required outside radius of the clevis end to not exceed the
above allowable stresses in either tear out or bearing if the clevis flanges are each 0.8 in thick.
Given:
Desired rod stress σrod  30 ksi
Nrods  10
L  30 ft
S sallow  20 ksi
Number of rods
Rod length
Clevis strength
d  0.75 in
Rod diameter
Young's modulus
Clevis flange thickness
6
E  30 10  psi
t  0.8 in
S ballow  40 ksi
Assumptions: The rods share the load equally, and there is one clevis for all ten rods.
Solution:
See Figures 4-12 and 4-13 in the text, Figure 4-19, and Mathcad file P0419.
A 
π d
2
2
A  0.442  in
1.
Calculate the cross-sectional area of one rod.
2.
Determine the force required to achieve the desired stress level in one rod.
F
σrod =
F  σrod  A
F  13.254 kip
A
3.
Determine the total force required to achieve the desired stress level in all rods.
Ftotal  Nrods F
4
Ftotal  132.5  kip
This force is transmitted through the clevis pin, which is in double shear.
4.
Calculate the minimum required clevis pin diameter for the allowable shear stress. Divide the load by 2
because of the double shear loading.
τpin =
Ftotal
2  Apin
Solving for the pin diameter
=
2  Ftotal
π d
2
= S sallow
d 
2  Ftotal
π S sallow
d  2.054  in
Round this up to the next higher decimal equivalent of a common fraction ( 2 1/8)
5.
d  2.125  in
Check the bearing stress in the clevis due to the pin on one side of the clevis.
Bearing stress area
Ab  d  t
Bearing force
Fb 
Bearing stress
σb 
Ftotal
2
Fb
Ab
2
Ab  1.700  in
Fb  66.268 kip
σb  39.0 ksi
Since this is less than S ballow, this pin diameter is acceptable.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6.
4-19-2
Determine the tearout stress in the clevis.
Shear area (see Figure 4-19)
2
Tearout length
2
Atear = 2  t R  ( 0.5 d )
Shear force
Ftear 
Ftotal
Ftear  66.268 kip
2
Shear stress and strength
R
d
τ=
Ftear
Atear
=
Ftear
2
2  t R  ( 0.5 d )
2
FIGURE 4-19
= S sallow
Tearout Diagram for Problem 4-19
2
Solving for the clevis radius, R
 Ftear 
2
R  
  ( 0.5 d)
2

t

S
sallow 

R  2.328  in
Round this up to the next higher decimal equivalent of a common fraction ( 2 3/8)
R  2.375  in
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-20-1
PROBLEM 4-20
Statement:
Repeat Problem 4-19 for 12 rods, each 1 cm in diameter and 10 m long. The desired rod stress is 20
MPa. The allowable normal stress in the clevis and pin is 280 MPa and their allowable shear stress
is 140 MPa. Each clevis flange is 2 cm wide.
Units:
kN  10  newton
Given:
Desired rod stress σrod  200  MPa
3
6
9
MPa  10  Pa
GPa  10  Pa
d  10 mm
Rod diameter
Number of rods
Nrods  12
Young's modulus
E  207  GPa
Rod length
Clevis strength
L  10 m
S sallow  140  MPa
Clevis flange thickness
t  20 mm
S ballow  280  MPa
Assumptions: The rods share the load equally, and there is one clevis for all twelve rods.
Solution:
See Figures 4-12 and 4-13 in the text, Figure 4-20, and Mathcad file P0420.
A 
π d
2
A  78.54 mm
1.
Calculate the cross-sectional area of one rod.
2.
Determine the force required to achieve the desired stress level in one rod.
σrod =
3.
F
F  σrod  A
A
4
2
F  15.708 kN
Determine the total force required to achieve the desired stress level in all rods.
Ftotal  Nrods F
Ftotal  188.5 kN
This force is transmitted through the clevis pin, which is in double shear.
4.
Calculate the minimum required clevis pin diameter for the allowable shear stress. Divide the load by 2 because
of the double shear loading.
τpin =
Ftotal
2  Apin
Solving for the pin diameter
=
2  Ftotal
π d
2
= S sallow
2  Ftotal
d 
π S sallow
d  30 mm
Round this up to the next higher even mm
5.
d  29.277 mm
Check the bearing stress in the clevis due to the pin on one side of the clevis.
Bearing stress area
Ab  d  t
Bearing force
Fb 
Bearing stress
σb 
Ftotal
2
Fb
Ab
Ab  600 mm
2
Fb  94.248 kN
σb  157.1 MPa
Since this is less than S ballow, this pin diameter is acceptable.
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P0420.xmcd
mechanical,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6.
4-20-2
Determine the tearout stress in the clevis.
Shear area (see Figure 4-19)
2
Tearout length
2
Atear = 2  t R  ( 0.5 d )
Shear force
Ftear 
Ftotal
Ftear  94.248 kN
2
Shear stress and strength
R
d
τ=
Ftear
Atear
=
Ftear
2
2  t R  ( 0.5 d )
2
= S sallow
FIGURE 4-20
Tearout Diagram for Problem 4-20
2
Solving for the clevis radius, R
 Ftear 
2
R  
  ( 0.5 d)
2

t

S
sallow 

Round this up to the next higher even mm
R  22.544 mm
R  24 mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-21-1
PROBLEM 4-21
Statement:
Figure P4-9 shows an automobile wheel with two common styles of lug wrench being used to
tighten the wheel nuts, a single-ended wrench in (a), and a double-ended wrench in (b). In each
case two hands are required to provide forces respectively at A and B as shown. The distance
between points A and B is 1 ft in both cases and the handle diameter is 0.625 in. The wheel nuts
require a torque of 70 ft-lb. Find the maximum principle stress and maximum deflection in each
wrench design.
Given:
Distance between A and B
d AB  1  ft
Tightening torque
Wrench diameter
T  70 ft  lbf
d  0.625  in
Assumptions: 1. The forces exerted by the user's hands lie in a plane through the wrench that is also parallel
to the plane of the wheel.
2. The applied torque is perpendicular to the plane of the forces.
3. By virtue of 1 and 2 above, this is a planar problem that can be described in a 2D FBD.
Solution:
See Figure 4-21 and Mathcad file P0421.
12" = dAB
1. In Problem 3-21 we found that for both cases
F
F  70 lbf
2. From examination of the FBDs, we see that, in
both cases, the arms are in bending and the stub
that holds the socket wrench is in pure torsion.
The maximum bending stress in the arm will occur
near the point where the arm transitions to the
stub. The stress state at this transition is very
complicated, but we can find the nominal bending
stress there by treating the arm as a cantilever
beam, fixed at the transition point. For both cases
the torque in the stub is the same.
T
F
(a) Single-ended Wrench
12" = dAB
F
6"
Case (a)
T
2. The bending moment at the transition is
F
Ma  F  d AB
(b) Double-ended Wrench
Ma  840  lbf  in
FIGURE 4-21
3. The tensile stress at this point is found from
Moment of inertia
4.
I 
Free Body Diagrams for Problem 4-21
π d
4
64
Dist to extreme fibre
c  0.5 d
Stress
σx 
M a c
I
4
I  0.00749  in
c  0.313  in
σx  35.05  ksi
There are no other stress components present at this point, so x is the maximum principle stress here and
σ1  σx
σ1  35.0 ksi
σ2  0  psi
σ3  0  psi
T  840  in lbf
5.
The torque in the stub is
6.
The shear stress at any point on the outside surface of the stub is found from
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
7.
Polar moment of inertia
J  2  I
Shear stress
τxy 
4
J  0.0150 in
Tc
τxy  17.52  ksi
J
There are no other stress components present along the outside surface of the stub, so
σ1  τxy
8.
4-21-2
σ1  17.5 ksi
σ2  0  psi
σ3  σ1
Thus, the maximum principle stress for case (a) is on the upper surface of the handle (arm) near the point
where it transitions to the stub.
There will be two deflection components that we can calculate separately and then add (superposition).
One will come from the bending of the arm and one will come from the twisting of the stub, projected out
to the end of the arm.
9.
Deflection of the arm due to bending only for a stub length of stub  3  in:
6
E  30 10  psi
Assuming that the wrenches are made from steel
6
G  11.7 10  psi
3
F  d AB
From Figure B-1(a), Appendix B,
yarm 
yarm  0.179  in
From equation (4.24), the
angular twist of the stub is
θstub 
The deflection at the end of
the arm due to the stub twist
is
ystub  d AB θstub
ystub  0.173  in
So, the total deflection is
ya  yarm  ystub
ya  0.352  in
3  E I
T  stub
θstub  0.014  rad
J G
Case (b)
Mb 
10. The bending moment at the transition is
F  d AB
2
Mb  420  lbf  in
11. The tensile stress at this point is found from
σx 
Stress
M b c
I
σx  17.52  ksi
12. There are no other stress components present at this point, so x is the maximum principle stress here and
σ1  σx
13. The torque in the stub is
14.
σ1  17.5 ksi
σ2  0  psi
σ3  0  psi
T  840  in lbf
The shear stress at any point on the outside surface of the stub is found from
Shear stress
τxy 
Tc
J
τxy  17.52  ksi
15. There are no other stress components present along the outside surface of the stub, so
σ1  τxy
σ1  17.5 ksi
σ2  0  psi
σ3  σ1
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-21-3
16. Thus, the maximum principle stress for case (b) is the same on the upper surface of the handle (arm) near the
point where it transitions to the stub, and on the outside surface of the stub.
There will be two deflection components that we can calculate separately and then add (superposition).
One will come from the bending of the arm and one will come from the twisting of the stub, projected out to
the end of the arm.
F   0.5 d AB
Deflection of the arm due to bending only:
From Figure B-1(a), Appendix B,
yarm 
From equation (4.24), the
angular twist of the stub is
θstub 
The deflection at the end of
the arm due to the stub twist
is
ystub 
So, the total deflection is
yb  yarm  ystub
3  E I
T  stub
J G
d AB
2
 θstub
3
yarm  0.022  in
θstub  0.014  rad
ystub  0.086  in
yb  0.109  in
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-22-1
PROBLEM 4-22
Statement:
A roller-blade skate is shown in Figure P4-10. The polyurethane wheels are 72 mm dia. The
skate-boot-foot combination weighs 2 kg. The effective "spring rate" of the person-skate
subsystem is 6000 N/m. The axles are 10-mm-dia steel pins in double shear. Find the stress in
the pins for a 100-kg person landing a 0.5-m jump on one foot. (a) Assume all 4 wheels land
simultaneously. (b) Assume that one wheel absorbs all the landing force.
Given:
Axle pin diameter
Solution:
See Figure P4-10 and Mathcad file P0422.
d  10 mm
Fa  897  N
Fb  3.59 kN
1.
From Problem 3-22, we have the forces for cases (a) and (b):
2.
In both cases, this is the force on one axle. The shear force will be one half of these forces because the pins
are in double shear.
Shear area
As 
π d
2
As  78.54  mm
4
2
Shear stress
Case (a) all wheels landing
τa 
Case (b) one wheel landing
τb 
Fa
2  As
Fb
2  As
τa  5.71 MPa
τb  22.9 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-23a-1
PROBLEM 4-23a
Statement:
A beam is supported and loaded as shown in Figure P4-11a. Find the reactions, maximum shear,
maximum moment, maximum slope, maximum bending stress, and maximum deflection for the
data given in row a from Table P4-2.
Given:
Beam length
L  1  m
Distance to distributed load
a  0.4 m
L
b
Distance to concentrated load b  0.6 m
w  200  N  m
Concentrated load
F  500  N
8
I  2.85 10
Moment of inertia
a
1
Distributed load magnitude
Distance to extreme fiber c  2.00 10
4
m
F
w
R2
R1
2
m
FIGURE 4-23A
Free Body Diagram for Problem 4-23
Solution:
1.
2.
See Figures 4-23 and Mathcad file P0423a.
The reactions, maximum shear and maximum moment were all found in Problem 3-23a. Those results are
summarized here.
Load function
q(x) = R1<x - 0>-1 - w<x - 0>0 + w<x - a>0 - F<x - b>-1 + R2<x - L>-1
Shear function
V(x) = R1<x - 0>0 - w<x - 0>1 + w<x - a>1 - F<x - b>0 + R2<x - L>0
Moment function
M(x) = R1<x - 0>1 - w<x - 0>2/2 + w<x - a>2/2 - F<x - b>1 + R2<x - L>1
Modulus of elasticity
E  207  GPa
Reactions
R1  264.0  N
Maximum shear
Vmax  316  N
Maximum moment
Mmax  126.4  N  m
R2  316.0  N
(negative, from x = b to x =L)
(at x = b)
Integrate the moment function, multiplying by 1/EI, to get the slope.
(x) = [R1<x>2/2 - w<x>3/6 + w<x - a>3/6 - F<x - b>2/2 + R2<x - L>2/2 + C3]/EI
3.
Integrate again to get the deflection.
y(x) = [R1<x>3/6 - w<x>4/24 + w<x - a>4/24 - F<x - b>3/6 + R2<x-L>3/6 + C3x +C4]/EI
4.
Evaluate C3 and C4
At x = 0 and x = L, y = 0, therefore, C4 = 0.
0=
R1
C3 
6
3
L 
w
24
4
L 
w
24
4
 ( L  a) 
F
6
3
 ( L  b )  C 3 L
 R1 3 w 4 w
F
4
3
L 
L 
 ( L  a)   ( L  b) 
L  6
24
24
6

1
 
2
C3  31.413 N  m
x  0  m 0.005  L  L
5.
Define the range for x
6.
For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
S ( x z)  if ( x  z 1 0 )
7.
Write the slope and deflection equations in Mathcad form, using the function S as a multiplying factor to
get the effect of the singularity functions. See Figure 4-23aB where these functions are plotted.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
θ ( x) 
y ( x) 
 R1
1
E I

2
E I
w
6
w
3
 S ( x 0  in)  x 
6

3
 S ( x a )  ( x  a ) 
 R
 2  S ( x L)  ( x  L) 2   F  S( x b)  ( x  b ) 2  C3
2
 2
 R1
1
2
 S ( x 0  in)  x 
4-23a-2

6
3
 S ( x 0  in)  x 
w
24
w
4
 S ( x 0  in)  x 



4
24
 R
 2  S( x L)  ( x  L) 3   F  S ( x b )  ( x  b) 3  C3 x
6
 6



θmax  θ ( L)
8. Maximum slope occurs at x = L

 S ( x a )  ( x  a ) 
θmax  0.335  deg
9. Maximum deflection occurs at x = c, where  = 0 and c < b.
θ0 =
 R1 2 w 3 w

3
 c   c   ( c  a )  C3 = 0
E I  2
6
6

1

w
a
Solving for c,
R1
A 
2
 3
6
B  3 
A  92.000 N
B 
c 
w
6
a
2
C  C3 
3
C  33.547 N  m
2
B  4 A  C
c  0.523  m
2 A
ymax  y ( c)
ymax  1.82 mm
SLOPE, radians
DEFECTION, mm
0.01
0
0.005
 0.5
y ( x)
0
mm
 0.005
 0.01
6
a
2
B  16.000 N  m
Substituting c into the deflection equation,
θ( x)
w
1
 1.5
0
0.2
0.4
0.6
0.8
1
2
0
0.2
0.4
0.6
x
x
m
m
0.8
1
FIGURE 4-23aB
Slope and Deflection Diagrams for Problem 4-23a
10. The maximum bending stress occurs at x = b, where the moment is a maximum. For
c  2.00 10
σmax 
2
m
Mmax c
I
c  20 mm
σmax  88.7 MPa
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-24a-1
PROBLEM 4-24a
Statement:
Given:
A beam is supported and loaded as shown in Figure P4-11b. Find the reactions, maximum
shear, maximum moment, maximum slope, maximum bending stress, and maximum deflection for
the data given in row a from Table P4-2.
Beam length
L  1  m
Distance to distributed load
a  0.4 m
L
a
F
Distance to concentrated load b  0.6 m
1
Distributed load magnitude
w  200  N  m
Concentrated load
F  500  N
8
I  2.85 10
Moment of inertia
Distance to extreme fiber c  2.00 10
Solution:
1.
2.
w
M1
4
m
R1
2
m
FIGURE 4-24A
Free Body Diagram for Problem 4-24
See Figures 4-24 and Mathcad file P0424a.
The reactions, maximum shear and maximum moment were all found in Problem 3-24a. Those results are
summarized here.
Load function
q(x) = -M1<x - 0>-2 + R1<x - 0>-1 - w<x - a>0 - F<x - L>-1
Shear function
V(x) = -M1<x - 0>-1 + R1<x - 0>0 - w<x - a>1 - F<x - L>0
Moment function
M(x) = -M1<x - 0>0 + R1<x - 0>1 - w<x - a>2/2 - F<x - L>1
Modulus of elasticity
E  207  GPa
Reactions
R1  620.0  N
Maximum shear
Vmax  620  N
Maximum moment
Mmax  584  N  m
M1  584.0  N  m
(positive, at x = 0)
(negative, at x = 0)
Integrate the moment function, multiplying by 1/EI, to get the slope.
(x) = [-M1<x-0>1 + R1<x - 0>2/2 - w<x - a>3/6 - F<x - L>2/2 + C3]/EI
3.
Integrate again to get the deflection.
y(x) = [-M1<x-0>2/2 + R1<x - 0>3/6 - w<x - a>4/24 - F<x - L>3/6 + C3x +C4]/EI
4.
Evaluate C3 and C4. At x = 0,  = 0 and y = 0, therefore, C3 = 0 and C4 = 0.
5.
Define the range for x
6.
For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
x  0  m 0.005  L  L
S ( x z)  if ( x  z 1 0 )
7.
Write the slope and deflection equations in Mathcad form, using the function S as a multiplying factor to get
the effect of the singularity functions. See Figure 4-24aB where these functions are plotted.
θ ( x) 
1
E I

 M1 S ( x 0  in)  x 
R1
2
 F
   S( x L)  ( x  L) 2
 2
2
 S ( x 0  in)  x 
w
6
3

 S ( x a )  ( x  a ) 



© 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.
MACHINE DESIGN - An Integrated Approach, 4th Ed.
y ( x) 
 M1
1
E I
 
2
 S ( x 0  in)  x 
R1
6
 2F
3
   S ( x L)  ( x  L)
 6
4-24a-2
3
 S ( x 0  in)  x 
w
24
4

 S ( x a )  ( x  a ) 



8. Maximum slope occurs at x = L
θmax  θ ( L)
θmax  2.73 deg
9. Maximum deflection occurs at x = L
ymax  y ( L)
ymax  32.2 mm
10. The maximum bending stress occurs at x = 0, where the moment is a maximum. For
σmax 
M1 c
σmax  410  MPa
I
SLOPE, radians
DEFLECTION, mm
0
0
 0.01
 10
 0.02
y ( x)
θ( x)
mm
 0.03
 20
 30
 0.04
 0.05
c  20 mm
0
0.2
0.4
0.6
0.8
1
 40
0
0.2
0.4
0.6
x
x
m
m
0.8
1
FIGURE 4-24aB
Slope and Deflection Diagrams for Problem 4-24a
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-25a-1
PROBLEM 4-25a
Statement:
A beam is supported and loaded as shown in Figure P4-11c. Find the reactions, maximum
shear, maximum moment, maximum slope, maximum bending stress, and maximum deflection for
the data given in row a from Table P4-2.
Given:
Beam length
L  1  m
Distance to distributed load
a  0.4 m
Distance to reaction load
b  0.6 m
L
b
a
Distributed load magnitude
w  200  N  m
Concentrated load
F  500  N
8
I  2.85 10
Moment of inertia
F
1
Distance to extreme fiber c  2.00 10
4
m
2
m
w
R2
R1
FIGURE 4-25A
Free Body Diagram for Problem 4-25
Solution:
1.
2.
See Figures 4-25 and Mathcad file P0425a.
The reactions, maximum shear and maximum moment were all found in Problem 3-25a. Those results are
summarized here.
Load function
q(x) = R1<x - 0>-1 - w<x - a>0 + R2<x - b>-1 - F<x - L>-1
Shear function
V(x) = R1<x - 0>0 - w<x - a>1 + R2<x - b>0 - F<x - L>0
Moment function
M(x) = R1<x - 0>1 - w<x - a>2/2 + R2<x - b>1 - F<x - L>1
Modulus of elasticity
E  207  GPa
Reactions
R1  353.3  N
Maximum shear
Vmax  580  N
Maximum moment
Mmax  216  N  m
R2  973.3  N
(positive, at x = b)
(negative, at x = b)
Integrate the moment function, multiplying by 1/EI, to get the slope.
(x) = [R1<x - 0>2/2 - w<x - a>3/6 + R2<x - b>2/2 - F<x - L>2/2 + C3]/EI
3.
Integrate again to get the deflection.
y(x) = [R1<x - 0>3/6 - w<x - a>4/24 + R2<x-b>3/6 - F<x - L>3/6 + C3x +C4]/EI
4.
Evaluate C3 and C4
At x = 0 and x = b, y = 0, therefore, C4 = 0.
0=
R1
6
3
b 
w
24
4
 ( b  a )  C3 b
1  R1 3
w
4
C3      b 
 ( b  a) 
b 6
24

2
C3  21.22  N  m
x  0  m 0.005  L  L
5.
Define the range for x
6.
For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
S ( x z)  if ( x  z 1 0 )
7.
Write the slope and deflection equations in Mathcad form, using the function S as a multiplying factor to get
the effect of the singularity functions. See Figure 4-25aB where these functions are plotted.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
θ ( x) 
y ( x) 
 R1
4-25a-2


E I 2
6
 R

 2  S ( x b )  ( x  b) 2   F  S( x L)  ( x  L) 2  C3 
2
 2

1

1

 R1
E I
2
 S ( x 0  in)  x 
w
3
 S ( x a )  ( x  a ) 


 R

 2  S( x b)  ( x  b ) 3   F  S ( x L)  ( x  L) 3  C3 x 
6
 6

6
3
 S ( x 0  in)  x 
w
24
4
 S ( x a )  ( x  a ) 
8. Maximum slope occurs at x = L
θmax  θ ( L)
θmax  0.823  deg
9. Maximum deflection occurs at x = L.
ymax  y ( L)
ymax  4.81 mm
10. The maximum bending stress occurs at x = b, where the moment is a maximum. For
σmax 
Mmax c
σmax  152  MPa
I
SLOPE, radians
DEFLECTION, mm
0.005
2
0
0
y ( x)
θ( x)  0.005
mm
 0.01
 0.015
c  20 mm
2
4
0
0.2
0.4
0.6
0.8
1
6
0
0.2
0.4
0.6
x
x
m
m
0.8
1
FIGURE 4-25aB
Slope and Deflection Diagrams for Problem 4-25a
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-26a-1
PROBLEM 4-26a
Statement:
A beam is supported and loaded as shown in Figure P4-11d. Find the reactions, maximum
shear, maximum moment, maximum slope, maximum bending stress, and maximum deflection for
the data given in row a from Table P4-2.
Given:
Beam length
L  1  m
a  0.4 m
Distance to R2
b  0.6 m
Distributed load magnitude
w  200  N  m
Concentrated load
F  500  N
b
8
Distance to extreme fiber c  2.00 10
4
m
2
m
F
w
R2
R1
R3
FIGURE 4-26A
Free Body Diagram for Problem 4-26
E  207  GPa
Modulus of elasticity
a
1
I  2.85 10
Moment of inertia
Solution:
L
Distance to distributed load
See Figures 4-26 and Mathcad file P0426a.
1. From inspection of Figure P4-11d, write the load function equation
q(x) = R1<x>-1 - F<x - a>-1 - w<x - a>0 + R2<x - b>-1 - R3<x - L>-1
2. Integrate this equation from - to x to obtain shear, V(x)
V(x) = R1<x>0 - F<x - a>0 - w<x - a>1 + R2<x - b>0 - R3<x - L>0
3. Integrate this equation from - to x to obtain moment, M(x)
M(x) = R1<x>1 - F<x - a>1 - w<x - a>2/2 + R2<x - b>1 - R3<x - L>1
4. Integrate the moment function, multiplying by 1/EI, to get the slope.
(x) = [R1<x>2/2 - F<x - a>2/2 - w<x - a>3/6 + R2<x - b>2/2 + R3<x - L>2/2 + C3]/EI
5. Integrate again to get the deflection.
y(x) = [R1<x>3/6 - F<x - a>3/6 - w<x - a>4/24 + R2<x - b>3/6 + R3<x - L>3/6 + C3x + C4]/EI
6. Evaluate R1, R2, R3, C3 and C4
At x = 0, x = b, and x = L; y = 0, therefore, C4 = 0.
At x = L+, V = M = 0
R1  100  N
Guess
R2  100  N
2
R3  100  N
C3  5  N  m
Given
R1
6
R1
6
b 
F
3
F
3
L 
6
6
3
 ( b  a) 
3
 ( L  a) 
w
24
w
24
4
3
 ( b  a )  C3 b = 0  N  m
4
 ( L  a) 
R2
6
3
3
 ( L  b )  C 3 L = 0  N  m
R1  F  w  ( L  a )  R2  R3 = 0  N
R 1 L  F  ( L  a ) 
w
2
2
 ( L  a )  R 2 ( L  b ) = 0  N  m
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-26a-2
 R1 
R 
 2   Find  R R R C 
1 2 3
3
 R3 
 
 C3 
R1  112.33 N
R2  559.17 N
2
R3  51.50  N
C3  5.607  N  m
x  0  in 0.002  L  L
7. Define the range for x
8. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
S ( x z)  if ( x  z 1 0 )
9. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get
the effect of the singularity functions.
V ( x)  R1 S ( x 0  in)  F  S ( x a )  w S ( x a )  ( x  a )  R2 S ( x b )  R3 S ( x L)
M ( x)  R1 S ( x 0  in)  x  F  S ( x a )  ( x  a ) 
 R2 S ( x b )  ( x  b )
w
2
2
 S ( x a )  ( x  a ) 
10. Plot the shear and moment diagrams.
SHEAR, N
V ( x)
N
MOMENT, N-m
200
60
0
35
M ( x)
 200
10
Nm
 400
 15
 600
 40
0
200
400
600
800
3
1 10
0
200
400
600
x
x
mm
mm
800
3
1 10
FIGURE 4-26aB
Shear and Moment Diagrams for Problem 4-26a
11. From the diagram, we see that maximum shear occurs at x = b -,
Vmax  V ( b  0.001  mm)
Vmax  428  N
12. The maximum moment occurs at x = a,
Mmax  M ( a )
Mmax  44.9 N  m
13. Write the slope and deflection equations in Mathcad form, using the function S as a multiplying factor to get t
effect of the singularity functions. See Figure 4-26aB where these functions are plotted.
© 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,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
θ ( x) 
y ( x) 
1
E I
1
E I
 R1

2
2
 S ( x 0  in)  x 
F
2
4-26a-3
2
 S ( x a )  ( x  a ) 
w
6
3
 R
R
 2  S ( x b )  ( x  b) 2  3  S ( x L)  ( x  L) 2  C3
2
 2
 R1

6
3
 S ( x 0  in)  x 
F
6
3
 S ( x a )  ( x  a ) 
w
24



4
 R
R
 2  S( x b)  ( x  b ) 3  3  S( x L)  ( x  L) 3  C3 x
6
 6
θmax  0.0576 deg
15. Maximum deflection occurs between x = 0 and x = a
ymax  0.200  mm



16. The maximum bending stress occurs at x = a, where the moment is a maximum. For
Mmax c
DEFLECTION, mm
0.1
0.1
0.05
0
θ( x)
y ( x)
0
mm
 0.05
 0.1
c  20 mm
σmax  31.5 MPa
I
SLOPE, deg.
deg

 S ( x a )  ( x  a ) 
14. Maximum slope occurs between x = a and x = b
σmax 

 S ( x a )  ( x  a ) 
 0.1
 0.2
0
0.2
0.4
0.6
0.8
1
 0.3
0
0.2
0.4
0.6
x
x
m
m
0.8
1
FIGURE 4-26aC
Slope and Deflection Diagrams for Problem 4-26a
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-27-1
PROBLEM 4-27
Statement:
A storage rack is to be designed to hold the paper roll of Problem 4-8 as shown in Figure P4-12.
Determine suitable values for dimensions a and b in the figure. Consider bending, shear, and
bearing stresses. Assume an allowable tensile/compressive stress of 100 MPa and an allowable
shear stress of 50 MPa for both stanchion and mandrel, which are steel. The mandrel is solid
and inserts halfway into the paper roll. Balance the design to use all of the material strength.
Calculate the deflection at the end of the roll.
Given:
Paper roll dimensions
Roll density
OD  1.50 m
Material properties
S y  100  MPa
ID  0.22 m
S ys  50 MPa
Lroll  3.23 m
E  207  GPa
3
ρ  984  kg m
Assumptions: 1. The paper roll's weight creates a concentrated load acting at the tip of the mandrel.
2. The mandrel's root in the stanchion experiences a distributed load over its length of
engagement
Solution:
See Figures 4-27 and Mathcad file P0427.
1. In Problem 3-27, we were concerned only with the
portion of the mandrel outside of the stanchion.
Therefore, we modeled it as a cantilever beam with a
shear and moment reaction at the stanchion.
Unfortunately, this tells us nothing about the stress or
force distributions in the portion of the mandrel that is
inside the stanchion. To do this we need to modify the
model by replacing the concentrated moment (and
possibly the concentrated shear force) with a force
system that will yield information about the stress
distribution in the mandrel on that portion that is inside
the stanchion. Figure 4-27A shows the FBD used in
Problem 3-27. Figure 4-27B is a simple model, but is not
representative of a built-in condition. It would be
appropriate if the hole in the stanchion did not fit tightly
around the mandrel. Figure 4-27C is an improvement
that will do for our analysis.
y
x
M1

4
π
2
2

 OD  ID  Lroll  ρ  g
Lm  0.5 Lroll
W  53.9 kN
Lm
R1
FIGURE 4-27A
Free Body Diagram for Problem 3-27
y
W
R1
x
2. Determine the weight of the roll and the length of the
mandrel.
W 
W
Lm
R2
FIGURE 4-27B
Simplified Free Body Diagram, not used
Lm  1.615  m
3. From inspection of Figure 4-27C, write the
load function equation
W
y
w
q(x) = -w<x>0 + w<x - b>0 + R<x - b>-1 - W<x - b -Lm>-1
a
x
4. Integrate this equation from - to x to obtain shear,
V(x)
V(x) = -w<x>1 + w<x - b>1 + R<x - b>0 - W<x - b -Lm>0
5. Integrate this equation from - to x to obtain
moment, M(x)
Lm
b
R
FIGURE 4-27C
Free Body Diagram used in Problem 4-27
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-27-2
M(x) = -(w/2)<x>2 + (w/2)<x - b>2 + R<x - b>1 - W<x - b -Lm>1
6. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = b + Lm,
where both are zero.
At x = (b + Lm)+ , V = M = 0
0 = w  b  Lm  w  Lm  R  W
R = W  w b
w
w
2 w
2
2 w
2
0 =    b  Lm   Lm  R Lm =    b  Lm   Lm  ( W  w b )  Lm
2
2
2
2
w=
2  W  Lm
b
2
Note that R is inversely proportional to b and w is inversly proportional to b 2.
7. To see the value of x at which the shear and moment are maximum, let
b  400  mm
w 
then
2  W  Lm
b
R  W  w b
and
2
L  b  Lm
x  0  mm 0.002  L  L
8. Define the range for x
9. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
S ( x z)  if ( x  z 1 0 )
10. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get
the effect of the singularity functions.
V ( x)  w S ( x 0  mm)  x  w S ( x b )  ( x  b )  R S ( x b )  W  S ( x L)
M ( x) 
w
2
2
 S ( x 0  mm)  x 
w
2
2
 S ( x b )  ( x  b )  R S ( x b )  ( x  b )  W  S ( x L)  ( x  L)
11. Plot the shear and moment diagrams.
Shear Diagram
Moment Diagram
200
50
0
V ( x)
kN
0
 200
M ( x)
kN  m
 400
 50
 600
 800
0
400
800
1200
1600
2000
 100
0
400
800
1200
x
x
mm
mm
1600
2000
FIGURE 4-27D
Shear and Moment Diagram Shapes for Problem 4-27
12. From Figure 4-27D, the maximum internal shear and moment occur at x = b and are
Vmax =
2  W  Lm
b
Mmax  W  Lm
Mmax  87.04  kN  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.
MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-27-3
13. The bending stress will be a maximum at the top or bottom of the mandrel at a section through x = b.
σmax =
Mmax a
where
2 I
I=
π a
4
so,
σmax =
64
32 Mmax
π a
1
 32 W  Lm 
 π S 
y 

Solving for a,
a 
Round this to
a  210  mm
3
= Sy
3
a  206.97 mm
14. Using this value of a and equation 4.15c, solve for the shear stress on the neutral axis at x = b.
τmax =
4  Vmax
3 A
=
8  W  Lm
 π a 2 
 b
3 
 4 
= S ys
8  W  Lm
Solving for b
b 
Round this to
b  134  mm
b  134.026  mm
 π a 2 
  Sys
3 
 4 
15. These are minimum values for a and b. Using them, check the bearing stress.
Magnitude of distributed load
w 
2  W  Lm
b
Bearing stress
σbear 
w  9695
2
w b
N
mm
σbear  46.2 MPa
a b
Since this is less than S y, the design is acceptable for a  210  mm and b  134  mm
16. Assume a cantilever beam loaded at the tip with load W and a mandrel diameter equal to a calculated above.
Moment of inertia
I 
π a
4
7
I  9.547  10  mm
64
4
3
Deflection at tip (Appendix B)
ymax  
W  Lm
3  E I
ymax  3.83 mm
This can be accomodated by the 220-mm inside diameter of the paper roll.
© 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,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-28-1
PROBLEM 4-28
Statement:
Figure P4-13 shows a forklift truck negotiating a 15 deg ramp to to drive onto a 4-ft-high loading
platform. The truck weighs 5 000 lb and has a 42-in wheelbase. Design two (one for each side)
1-ft-wide ramps of steel to have no more than 1-in deflection in the worst case of loading as the
truck travels up them. Minimize the weight of the ramps by using a sensible cross-sectional
geometry.
Given:
Ramp angle
Platform height
θ  15 deg
Truck weight
Truck wheelbase
W  5000 lbf
Lt  42 in
h  4  ft
Ramp width
Allowable deflection
w  12 in
δmax  1.0 in
Young's modulus
E  30 10  psi
6
Assumptions: 1. The worst case is when the truck CG is located at the center of the beam's span.
2. Use a coordinate frame that has the x-axis along the long axis of the beam.
3. Ignore traction forces and the weight components along the x-axis of the beam.
4. There are two ramps, one for each side of the forklift.
Solution:
See Figure 4-28 and Mathcad file P0428.
L
b
a
CG a
y
CG b
R1

Fa
Wa
Fb
x
Wb
R2
FIGURE 4-28A
Dimensions and Free Body Diagram for Problem 4-28
1. Determine the length of the beam between supports and the distances a and b for the worst-case loading.
h
Length of beam
L 
From Problem 3-28,
a  5.061  ft
sin( θ )
L  15.455 ft
b  8.561  ft
2. The load distribution of the wheels on a single ramp is given in Problem 3-28 as
Fa  575.0  lbf
Fb  1839.9 lbf
3. From inspection of Figure 4-28A, write the load function equation
q(x) = R1<x - 0>-1 - Fa<x - a>-1 - Fb<x - b>-1 + R2<x - L>-1
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-28-2
4. Integrate this equation from - to x to obtain shear, V(x)
V(x) = R1<x - 0>0 - Fa<x - a>0 - Fb<x - b>0 + R2<x - L>0
5. Integrate this equation from - to x to obtain moment, M(x)
M(x) = R1<x - 0>1 - Fa<x - a>1 - Fb<x - b>1 + R2<x - L>1
R1  1207.4 lbf
6. The reactions are given in Problem 3-28 as
R2  1207.4 lbf
7. Integrate the moment function, multiplying by 1/EI, to get the slope.
(x) = [R1<x>2/2 - Fa<x - a>2/2 - Fb<x - b>2/2 + R2<x - L>2/2 + C3]/EI
8. Integrate again to get the deflection.
y(x) = [R1<x>3/6 - Fa<x - a>3/6 - Fb<x - b>3/6 + R2<x-L>3/6 + C3x +C4]/EI
9. Evaluate C3 and C4
At x = 0 and x = L, y = 0, therefore, C4 = 0.
3
3
3
0 = R1 L  Fa ( L  a )  Fb ( L  b )  6  C3 L
1
C3 
 R1 L  Fa ( L  a )  Fb ( L  b )
3
6 L
3
3
6
2
C3  4.983  10  lbf  in

x  0  m 0.005  L  L
8. Define the range for x
9. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
S ( x z)  if ( x  z 1 0 )
10. Write the slope and deflection equations in Mathcad form, using the function S as a multiplying factor to get
the effect of the singularity functions. Use an assumed value of I so that the value of x that corresponds to ymax
4
can be found. Let I  10 in
θ ( x) 
y ( x) 
1
E I
1
E I
 R1

2
2
 S ( x 0  m)  x 
Fa
2
2
 S ( x a )  ( x  a ) 
 R
 2  S ( x L)  ( x  L) 2  C3
 2
 R1

6
3
 S ( x 0  m)  x 
Fa
6
3
 S ( x a )  ( x  a ) 
 R
 2  S( x L)  ( x  L) 3  C3 x
 6
Fb
2
Fb
6
2

 S ( x b )  ( x  b ) 



3

 S ( x b )  ( x  b ) 



11. Plot the shear and moment diagrams using the assumed value of I.
© 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,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-28-3
SLOPE, radians
DEFLECTION, in
0.02
0
0.01
 0.5
θ( x)
y ( x)
0
in
1
 0.01
 0.02
0
4
8
12
 1.5
16
0
4
8
x
x
ft
ft
12
16
FIGURE 4-28B
Slope and Deflection Diagrams for Problem 4-28, Using an Assumed Value for I
12. Maximum deflection occurs at x = c, where  = 0 and c < b.
θ0 =
 R1 2 Fa

2
c 
 ( c  a )  C3 = 0
E I  2
2

1

Solving for c,
A 
R1
2

2
Fa
B  a  Fa
2
4
A  316.200  lbf
c 
C  C3 
a  Fa
2
6
B  3.492  10  lbf  in
2
C  6.043  10  in  lbf
2
B 
B  4 A  C
c  7.804  ft
2 A
13. The maximum deflection occurs at x = c and is
ymax =
1
E I
 R 1 c 3

 6

Fa
6

 ( c  a )  C3 c = δmax
3

Solving for I
I 
1
E δmax
 R 1 c 3


6

Fa
6

 ( c  a )  C3 c
3
4
I  10.159 in

This is the minimum allowable value of the moment of inertia.
14. Assume a channel section such as that shown in Figure 4-28C. To keep it simple, let the thickness of
the flanges and web be the same. Choose 3/8-in thick plate, which is readily available. Then,
t  0.375  in
15. The cross-sectional area of the ramp is
16. The distance to the CG is
cg( h ) 
A ( h )  w t  2  t ( h  t)
1
A (h)
 w t 2

 2
2
 t h  t
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,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-28-4
17. The moments of inertia of the web and a flange are
w t
3
Iweb( h ) 
 w t  cg( h ) 
12

Ifl ( h ) 
t ( h  t)
12
3


2
t
 h  t  cg( h ) 

Flange
2
Web
h  t
2
t

2 
18. Using the known moment of inertia, solve
for the unknown flange height, h. Guess
h  1  in
Given
I = Iweb( h )  2  Ifl ( h )
h  Find ( h )
Round this up to
h
h  3.988  in
h  4.00 in
w
FIGURE 4-28C
Channel Section for Problem 4-28
19. Summarizing, the ramp design dimensions are:
Length
L  185.5  in
Flange height
h  4.00 in
Shape
channel
Width
w  12.00  in
Thickness
t  0.375  in
Material
steel
© 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,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-29a-1
PROBLEM 4-29a
Statement:
Find the spring rate of the beam in Problem 4-23 at the applied concentrated load for row a in
Table P4-2.
Given:
Beam length
L  1  m
Distance to distributed load
a  0.4 m
L
b
a
Distance to concentrated load b  0.6 m
Solution:
1
Distributed load magnitude
w  200  N  m
Concentrated load
Fb  500  N
8
Moment of inertia
I  2.85 10
Modulus of elasticity
E  207  GPa
F
w
R2
R1
4
m
FIGURE 4-29
Free Body Diagram for Problem 4-23
See Figure 4-29 and Mathcad file P0429a.
1. The deflection equation was found in Problem 4-23. Those results are summarized here.
Load function
q(x) = R1<x - 0>-1 - w<x - 0>0 + w<x - a>0 - F<x - b>-1 + R2<x - L>-1
Shear function
V(x) = R1<x - 0>0 - w<x - 0>1 + w<x - a>1 - F<x - b>0 + R2<x - L>0
Moment function
M(x) = R1<x - 0>1 - w<x - 0>2/2 + w<x - a>2/2 - F<x - b>1 + R2<x - L>1
Slope function
(x) = [R1<x>2/2 - w<x>3/6 + w<x - a>3/6 - F<x - b>2/2 + R2<x - L>2/2 + C3]/EI
Deflection function
y(x) = [R1<x>3/6 - w<x>4/24 + w<x - a>4/24 - F<x - b>3/6 + R2<x-L>3/6 + C3x +C4]/EI
2. To determine the stiffness under the load F we will need to find the incremental beam deflection due to F
alone. The procedure will be to find the deflection at x = b when F = 0, and then find it when Fb  500  N . The
stiffness will then be the force divided by the incremental deflection.
3. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
S ( x z)  if ( x  z 1 0 )
4. Write the reactions (from Problem 3-23), integration constant, and deflection (from problem 4-23) equations in
Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions.
R1( F ) 
w
2
L 
F
L
 ( L  b) 
w
2 L
 ( L  a)
2
R2( F )  w a  F  R1( F )
C3( F ) 
y ( x F ) 
 R1( F ) 3 w 4 w
F
4
3
L 
L 
 ( L  a)   ( L  b) 
6
L 
24
24
6

1
 
1
E I
 R1( F )

6
3
 S ( x 0  in)  x 
w
24
w
4
 S ( x 0  in)  x 
24
4

 S ( x a )  ( x  a ) 
 R (F )
 2  S( x L)  ( x  L) 3   F  S ( x b )  ( x  b) 3  C3( F )  x
6
 6



5. The deflection at x = b for F  0  N is
y0  y ( b F )
y0  0.137  mm
6. The deflection at x = b for F  Fb is
yF  y ( b F )
yF  1.765  mm
Δy  yF  y0
Δy  1.627  mm
7. The deflection due to F alone is
8. The stiffness of the beam under the load F at x = b is
k 
F
Δy
k  307 
N
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,
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Pearson Education, Inc., Upper Saddle River, NJ 07458.
MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-30a-1
PROBLEM 4-30a
Statement:
Find the spring rate of the beam in Problem 4-24 at the applied concentrated load for row a in
Table P4-2.
Given:
Beam length
Solution:
L  1  m
L
Distance to distributed load
a  0.4 m
Distributed load magnitude
w  200  N  m
Concentrated load
FL  500  N
a
1
8
Moment of inertia
I  2.85 10
Modulus of elasticity
E  207  GPa
F
w
4
m
M1
R1
See Figure 4-30 and Mathcad file P0430a.
FIGURE 4-30
1. The deflection equation was found in Problem 4-24.
Those results are summarized here.
Free Body Diagram for Problem 4-24
Load function
q(x) = -M1<x - 0>-2 + R1<x - 0>-1 - w<x - a>0 - F<x - L>-1
Shear function
V(x) = -M1<x - 0>-1 + R1<x - 0>0 - w<x - a>1 - F<x - L>0
Moment function
M(x) = -M1<x - 0>0 + R1<x - 0>1 - w<x - a>2/2 - F<x - L>1
Slope function
(x) = [-M1<x-0>1 + R1<x - 0>2/2 - w<x - a>3/6 - F<x - L>2/2 + C3]/EI
Deflection function
y(x) = [-M1<x-0>2/2 + R1<x - 0>3/6 - w<x - a>4/24 - F<x - L>3/6 + C3x +C4]/EI
2. To determine the stiffness under the load F we will need to find the incremental beam deflection due to
F alone. The procedure will be to find the deflection at x = L when F = 0, and then find it when FL  500  N
. The stiffness will then be the force divided by the incremental deflection.
3. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
S ( x z)  if ( x  z 1 0 )
4. Write the reaction (from Problem 3-24) and deflection (from problem 4-24) equations in Mathcad form, using
the function S as a multiplying factor to get the effect of the singularity functions.
R1( F )  w ( L  a )  F
M1( F ) 
y ( x F ) 
w
2
2
 ( L  a )  R 1( F )  L
1
E I
 M1( F )
 
2
F
2
 S ( x 0  in)  x 

   S( x L)  ( x  L) 3
 6
R1( F )
6
3
 S ( x 0  in)  x 
w
24
4



5. The deflection at x = L for F  0  N is
y0  y ( L F )
y0  3.912  mm
6. The deflection at x = L for F  FL is
yF  y ( L F )
yF  32.163 mm
Δy  yF  y0
Δy  28.251 mm
7. The deflection due to F alone is
8. The stiffness of the beam under the load F at x = L is
k 
F
Δy

 S ( x a )  ( x  a ) 
k  17.7
N
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.
MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-31a-1
PROBLEM 4-31a
Statement:
Find the spring rate of the beam in Problem 4-25 at the applied concentrated load for row a in
Table P4-2.
Given:
Beam length
L  1  m
Distance to distributed load
a  0.4 m
L
b
a
Distance to concentrated load b  0.6 m
Solution:
1
Distributed load magnitude
w  200  N  m
Concentrated load
FL  500  N
8
Moment of inertia
I  2.85 10
Modulus of elasticity
E  207  GPa
F
w
R2
R1
4
m
FIGURE 4-31
Free Body Diagram for Problem 4-25
See Figure 4-31 and Mathcad file P0431a.
1. The deflection equation was found in Problem 4-25. Those results are summarized here.
Load function
q(x) = R1<x - 0>-1 - w<x - a>0 + R2<x - b>-1 - F<x - L>-1
Shear function
V(x) = R1<x - 0>0 - w<x - a>1 + R2<x - b>0 - F<x - L>0
Moment function
M(x) = R1<x - 0>1 - w<x - a>2/2 + R2<x - b>1 - F<x - L>1
Slope function
(x) = [R1<x - 0>2/2 - w<x - a>3/6 + R2<x - b>2/2 - F<x - L>2/2 + C3]/EI
Deflection function
y(x) = [R1<x - 0>3/6 - w<x - a>4/24 + R2<x-b>3/6 - F<x - L>3/6 + C3x +C4]/EI
2. To determine the stiffness under the load F we will need to find the incremental beam deflection due to
F alone. The procedure will be to find the deflection at x = L when F = 0, and then find it when
FL  500  N . The stiffness will then be the force divided by the incremental deflection.
3. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
S ( x z)  if ( x  z 1 0 )
4. Write the reactions (from Problem 3-25), integration constant, and deflection (from problem 4-25) equations in
Mathcad form, using the function S as a multiplying factor to get the effect of the singularity functions.
R1( F ) 
w
2
   ( L  a )  F  ( L  b )  w ( L  a )  ( L  b )
b 2

1
R2( F )  w ( L  a )  F  R1( F )
1  R1( F ) 3
w
4
C3( F )    
b 
 ( b  a) 
b  6
24

y ( x F ) 


6
E I
24
 R (F )

 2  S( x b)  ( x  b ) 3   F  S ( x L)  ( x  L) 3  C3( F )  x 
6
 6

1
 R1( F )

3
 S ( x 0  in)  x 
w
4
 S ( x a )  ( x  a ) 
5. The deflection at x = L for F  0  N is
y0  y ( L F )
y0  0.288  mm
6. The deflection at x = L for F  FL is
yF  y ( L F )
yF  4.808  mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
7. The deflection due to F alone is
4-31a-2
Δy  yF  y0
8. The stiffness of the beam under the load F at x = L is
Δy  4.52 mm
k 
F
Δy
k  111 
N
mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-32a-1
PROBLEM 4-32a
Statement:
Find the spring rate of the beam in Problem 4-26 at the applied concentrated load for row a in
Table P4-2.
Given:
Beam length
L  1  m
Distance to distributed load
a  0.4 m
L
b
a
Distance to concentrated load b  0.6 m
Solution:
F
w
1
Distributed load magnitude
w  200  N  m
Concentrated load
Fa  500  N
8
Moment of inertia
I  2.85 10
Modulus of elasticity
E  207  GPa
R2
R1
R3
4
m
FIGURE 4-32
Free Body Diagram for Problem 4-26
See Figure 4-32 and Mathcad file P0432a.
1. The deflection equation was found in Problem 4-26. Those results are summarized here.
Load function
q(x) = R1<x>-1 - F<x - a>-1 - w<x - a>0 + R2<x - b>-1 - R3<x - L>-1
Shear function
V(x) = R1<x>0 - F<x - a>0 - w<x - a>1 + R2<x - b>0 - R3<x - L>0
Moment function
M(x) = R1<x>1 - F<x - a>1 - w<x - a>2/2 + R2<x - b>1 - R3<x - L>1
Slope function
(x) = [R1<x>2/2 - F<x - a>2/2 - w<x - a>3/6 + R2<x - b>2/2 + R3<x - L>2/2 + C3]/EI
Deflection function y(x) = [R1<x>3/6 - F<x - a>3/6 - w<x - a>4/24 + R2<x - b>3/6 + R3<x - L>3/6 + C3x + C4]/EI
2. To determine the stiffness under the load F we will need to find the incremental beam deflection due to
F alone. The procedure will be to find the deflection at x = a when F = 0, and then find it when Fa  500  N
. The stiffness will then be the force divided by the incremental deflection.
3. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
S ( x z)  if ( x  z 1 0 )
4. Write the reactions, integration constant, and deflection (from problem 4-26) equations in Mathcad form,
using the function S as a multiplying factor to get the effect of the singularity functions.
Let
f1 ( F ) 
F
6
3
 ( b  a) 
w
f3 ( F )  F  ( L  a ) 
2
w
24
 ( b  a)
 ( L  a)
4
f2 ( F ) 
F
6
3
 ( L  a) 
w
24
 ( L  a)
4
2
then
R1( F ) 
R2( F ) 
 L
3
L b  ( L  b )
1
( L  b)

b
 f1 ( F )  f2 ( F ) 
  f3 ( F )  L R1( F ) 
R3( F )  F  w ( L  a )  R1( F )  R2( F )
C3( F ) 
1
b
 f1 ( F ) 
b
( L  b)
6
2

 f3 ( F )

R1 Fa  112.333  N
R2 Fa  559.167  N
R3 Fa  51.500 N
2
6
 R 1( F )
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
y ( x F ) 
1
E I
 R1( F )

6
3
 S ( x 0  in)  x 
F
6
4-32a-2
3
 S ( x a )  ( x  a ) 
w
24
4
 R (F )
R (F )
 2  S( x b)  ( x  b ) 3  3  S ( x L)  ( x  L) 3  C3( F )  x
6
 6
5. The deflection at x = a for F  0  N is
y0  y ( a F )
y0  0.00126  mm
6. The deflection at x = a for F  Fa is
yF  y ( a F )
yF  0.177  mm
Δy  yF  y0
Δy  0.176  mm
7. The deflection due to F alone is
8. The stiffness of the beam under the load F at x = a is
k 
F
Δy

 S ( x a )  ( x  a ) 



k  2844
N
mm
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-33a-1
PROBLEM 4-33a
Statement:
For the bracket shown in Figure P4-14 and the data in row a of Table P4-3, determine the
bending stress at point A and the shear stress due to transverse loading at point B. Also the
torsional shear stress at both points. Then determine the principal stresses at points A and B.
Given:
Tube length
L  100  mm
F
y
Arm length
a  400  mm
Arm thickness
t  10 mm
Arm depth
h  20 mm
Applied force
F  50 N
Tube OD
OD  20 mm
A
B
T
T
x
M
L
R
Tube ID
ID  14 mm
Modulus of elasticity
E  207  GPa
FIGURE 4-33
Free Body Diagram of Tube for Problem 4-33
Solution:
See Figure 4-33 and Mathcad file P0433a.
1. Determine the bending stress at point A. From the FBD of the tube in Figure 4-33 we see that
Reaction force
R  F
R  50.0 N
Reaction moment
M  F  L
M  5.00 N  m
Distance from NA
to outside of tube
ct  0.5 OD
ct  10.0 mm
Moment of inertia
It 
Bending stress
at point A
σxA 

64
π
4
 OD  ID
4

M  ct
It  5968 mm
4
σxA  8.38 MPa
It
2. Determine the shear stress due to transverse loading at B.
Cross-section area
A 
π
4

2
 OD  ID
Maximum shear
V  R
Maximum shear stress
(Equation 4.15d)
τVmax  2 
2

V
A  160.2  mm
2
τVmax  0.624  MPa
A
3. Determine the torsional shear stress at both points. Using equation 4.23b and the FBD above
Torque on tube
T  F  a
Polar moment of
inertia
J 
Maximum torsional
stress at surface
τTmax 
T  20.0 N  m

32
π
4
 OD  ID
T  ct
J
4

J  11936  mm
4
τTmax  16.76  MPa
4. Determine the principal stress at point A.
Stress components
σxA  8.378  MPa
σzA  0  MPa
τxz  τTmax
τxz  16.76  MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-33a-2
Principal stresses
σ1 
σxA  σzA
2
2
 σxA  σzA 
2
 
  τxz
2


σ1  21.46  MPa
2
 σxA  σzA 
2
 
  τxz
2


σ3  13.08  MPa
σ2  0  MPa
σ3 
τ13 
σxA  σzA
2
σ1  σ3
τ13  17.27  MPa
2
5. Determine the principal stress at point B.
Stress components
σxB  0  MPa
σyB  0  MPa
τxy  τTmax  τVmax
τxy  16.13  MPa
Principal stresses
σ1 
σxB  σyB
2
2
 σxB  σyB 
2
 
  τxy
2


σ1  16.13  MPa
2
 σxB  σyB 
2
 
  τxy
2


σ3  16.13  MPa
σ2  0  MPa
σ3 
τ13 
σxB  σyB
2
σ1  σ3
2
τ13  16.13  MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-34a-1
PROBLEM 4-34a
Statement:
For the bracket shown in Figure P4-14 and the data in row a of Table P4-3, determine the
deflection at load F.
Given:
Tube length
Arm length
Arm thickness
Arm depth
Solution:
See Figure 4-34 and Mathcad file P0434a.
L  100  mm
a  400  mm
t  10 mm
h  20 mm
Applied force
Tube OD
Tube ID
Modulus of elasticity
Modulus of rigidity
F  50 N
OD  20 mm
ID  14 mm
E  207  GPa
G  80.8 GPa
1.
The deflection at load F can be determined by superimposing the rigid-body deflection of the arm due to the
twisting of the tube with the beam deflection of the tube and the arm alone.
2.
Determine the rigid-body deflection due to twisting of the tube. Refering to Figure 4-34, the torque in the
tube is
Torque on tube
T  F  a
Polar moment of inertia
Jt 
Tube angle of twist
θ 
T  20.0 N  m

32
π
4
 OD  ID
4

Jt  11936  mm
TL
4
θ  2.07368  10
J t G
3
 rad
θ  0.119  deg
Deflection at F due to 
3.
δθ  a  θ
δθ  0.829  mm
Determine the rigid-body deflection due to bending of the tube.
It 
Moment of inertia
Deflection of tube
end and arm end
(see Appendix B)
Jt
It  5968 mm
2
4
3
δtb 
F L
δtb  0.013  mm
3  E It
F
F
y
a
y
A
B
T
T
x
z
M
h
T
L
F
R
FIGURE 4-34
Free Body Diagrams of Tube and Arm for Problem 4-34
4.
Determine the beam bending of arm alone.
Moment of inertia
Deflection at F
5.
Ia 
δa 
t h
3
Ia  6667 mm
12
F a
4
3
δa  0.773  mm
3  E Ia
Determine the total deflection by superposition.
δtot  δθ  δtb  δa
δtot  1.616  mm
downward
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-35a-1
PROBLEM 4-35a
Statement:
For the bracket shown in Figure P4-14 and the data in row a of Table P4-3, determine the spring
rate of the tube in bending, the spring rate of the arm in bending, and the spring rate of the tube
in torsion. Combine these into an overall spring rate in terms of the force F and the linear
deflection at F.
Given:
Tube length
Arm length
Arm thickness
Arm depth
Solution:
See Figure 4-35 and Mathcad file P0435a.
1.
2.
L  100  mm
a  400  mm
t  10 mm
h  20 mm
Applied force
Tube OD
Tube ID
Modulus of elasticity
Modulus of rigidity
F  50 N
OD  20 mm
ID  14 mm
E  207  GPa
G  80.8 GPa
Determine the spring rate due to bending of the tube.

64
π
Moment of inertia
It 
Deflection of tube
end and arm end
(see Appendix B)
δtb 
Spring rate due to
bending in tube
ktb 
4
4
 OD  ID

It  5968 mm
4
3
F L
δtb  0.013  mm
3  E It
F
ktb  3706
δtb
N
mm
Determine the spring rate due to beam bending of arm alone.
t h
Ia 
Moment of inertia
3
Ia  6667 mm
12
Deflection at F
δa 
Spring rate due to
bending in arm
ka 
F a
4
3
δa  0.773  mm
3  E Ia
F
ka  64.7
δa
N
mm
F
F
y
a
y
A
B
T
T
x
z
M
h
T
L
F
R
FIGURE 4-35
Free Body Diagrams of Tube and Arm for Problem 4-35
3.
Determine the spring rate of the tube in torsion. Refering to Figure 4-35, the torque in the tube is
Torque on tube
Polar moment of inertia
T  F  a
Jt 

32
π
4
 OD  ID
4

T  20.0 N  m
Jt  11936  mm
4
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
Tube angle of twist
θ 
4-35a-2
TL
θ  2.07368  10
J t G
3
 rad
θ  0.119  deg
4.
Deflection at F due to q
δθ  a  θ
Spring rate due to
torsion in tube
kθ 
F
kθ ktb ka
1
koa
=
1
kθ

1
ktb

δtot 
F
koa
N
mm
1
ka
koa  30.9
ktb ka  kθ ka  kθ ktb
Checking,
kθ  60.28 
δθ
Determine the overall spring rate. The springs are in series, thus
koa 
δθ  0.829  mm
N
mm
δtot  1.616  mm
which is the same total deflection gotten in Problem 4-34.
© 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,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-36a-1
PROBLEM 4-36a
Statement:
For the bracket shown in Figure P4-14 and the data in row a of Table P4-3, redo Problem 4-33
considering the stress concentration at points A and B. Assume a stress concentration factor
of 2.5 in both bending and torsion.
Given:
Tube length
L  100  mm
Arm length
a  400  mm
Arm thickness
t  10 mm
Arm depth
h  20 mm
Applied force
F  50 N
Tube OD
OD  20 mm
Tube ID
ID  14 mm
Modulus of elasticity
E  207  GPa
Stress-concentration
factors
Ktb  2.5
Solution:
1.
2.
3.
4.
F
y
A
B
T
T
x
M
L
R
FIGURE 4-36
Free Body Diagram of Tube for Problem 4-36
Kts  2.5
See Figure 4-36 and Mathcad file P0436a.
Determine the bending stress at point A. From the FBD of the tube in Figure 4-36 we see that
Reaction force
R  F
R  50.0 N
Reaction moment
M  F  L
M  5.00 N  m
Distance from NA
to outside of tube
ct  0.5 OD
ct  10.0 mm
Moment of inertia
It 
Bending stress
at point A
σxA  Ktb

64
π
4
 OD  ID
4

M  ct
It  5968 mm
4
σxA  20.94  MPa
It
Determine the shear stress due to transverse loading at B.

4
π
Cross-section area
A 
Maximum shear
V  R
Maximum shear stress
(Equation 4.15d)
τVmax  2 
2
 OD  ID
2

V
A  160.2  mm
2
τVmax  0.624  MPa
A
Determine the torsional shear stress at both points. Using equation 4.23b and the FBD above
Torque on tube
T  F  a
Polar moment of
inertia
J 
Maximum torsional
stress at surface
τTmax  Kts

32
π
T  20.0 N  m
4
 OD  ID
T  ct
J
4

J  11936  mm
4
τTmax  41.89  MPa
Determine the principal stress at point A.
Stress components
σxA  20.944 MPa
σzA  0  MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
τxz  τTmax
4-36a-2
τxz  41.89  MPa
Principal stresses
σ1 
σxA  σzA
2
2
 σxA  σzA 
2
 
  τxz
2


σ1  53.6 MPa
2
 σxA  σzA 
2
 
  τxz
2


σ3  32.71  MPa
σ2  0  MPa
σ3 
τ13 
5.
σxA  σzA
2
σ1  σ3
τ13  43.18  MPa
2
Determine the principal stress at point B.
Stress components
σxB  0  MPa
σyB  0  MPa
τxy  τTmax  τVmax
τxy  41.26  MPa
Principal stresses
σ1 
σxB  σyB
2
2
 σxB  σyB 
2
 
  τxy
2


σ1  41.26  MPa
2
 σxB  σyB 
2
 
  τxy
2


σ3  41.26  MPa
σ2  0  MPa
σ3 
τ13 
σxB  σyB
2
σ1  σ3
2
τ13  41.26  MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-37-1
PROBLEM 4-37
Statement:
Given:
A semicircular, curved beam as shown in Figure 4-37 has the dimensions given below. For the
load pair applied along the diameter and given below, find the eccentricity of the neutral axis
and the stress at the inner and outer fibers.
Outside diameter
od  150  mm
Inside diameter
id  100  mm
Width of beam
w  25 mm
Load
F  14 kN
w
F
od
Solution:
id
See Figure 4-37 and Mathcad file P0437.
F
1. Calculate the section depth, area, inside radius and
outside radus.
Section depth
h 
od  id
2
Area of section
A  h  w
Centroid radius
rc 
Inside and outside
radii of section
(a) Entire Beam
h  25 mm
A  625  mm
od  id
2
rc  62.5 mm
4
ri  rc  0.5 h
ri  50 mm
ro  rc  0.5 h
ro  75 mm
F
M
F
rc
(b) Critical Section
2. The critical section is the one that is along the horizontal
centerline. There, the bending moment is
Bending moment
3.
Free Body Diagrams for Problem 4-37
M  0.875  kN  m
Use the equation in the footnote of the text to calculate the radius of the neutral axis.
Radius of neutral axis
4.
M  F  rc
FIGURE 4-37
rn 
ro  ri
rn  61.658 mm
 ro 
ln 
 ri 
Calculate the eccentricty and the distances from the neutral axis to the extreme fibers.
Eccentricity
e  rc  rn
e  0.8424 mm
Distances from neutral
axis to extreme fibers
ci  rn  ri
ci  11.66  mm
co  ro  rn
co  13.34  mm
Stresses at inner and
outer radii
σi 
M

ci
e A ri

F
A
 M co  F
 
 e A ro  A
σo  
σi  409.9  MPa
σo  273.2  MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-38-1
PROBLEM 4-38
Statement:
Design a solid, straight, steel torsion bar to have a spring rate of 10 000 in-lb per radian per foot
of length. Compare designs of solid round and solid square cross-sections. Which is more
efficient in terms of material use?
Given:
Length of rod
L  12 in
Spring rate
k  10000 
Solution:
1.
Modulus of rigidity
6
G  11.7 10  psi
in lbf
rad
See Mathcad file P0438.
Determine the rod diameter and volume for a round rod.
Spring rate
k=
J G
J =
L
π d
4
32
1
2.
Rod diameter
d 
Volume of rod
V 
 32 L k 
 π G 


π d
4
4
d  0.569  in
2
3
L
V  3.046  in
Determine the rod width and volume for a square rod using equation 4.26b and Table 4-6.
Spring rate
k=
K G
K = 2.25 a
L
4
1
3.
 L k 
 2.25 G 


Rod half-width
a 
Volume of rod
V  ( 2  a )  L
2
4
a  0.260  in
2  a  0.520  in
3
V  3.241  in
Even though the square rod width is less than the round rod diameter, it takes slightly more material when a
square rod is used than when a round rod is used. Thus, the round rod is more efficient.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-39-1
PROBLEM 4-39
Statement:
Design a 1-ft-long steel, end-loaded cantilever spring for a spring rate of 10 000 lb/in. Compare
designs of solid round and solid square cross-sections. Which is more efficient in terms of
material use?
Given:
Length of rod
L  12 in
Spring rate
k  10000 
Solution:
1.
lbf
in
Determine the rod diameter and volume for a round rod.
k=
3  E I
I=
3
L
π d
4
64
1
Rod diameter
 64 L3 k 

d  
 3  π E 
Volume of rod
V 
π d
4
d  1.406  in
2
4
3
L
V  18.64  in
Determine the rod width and volume for a square rod using equation 4.26b and Table 4-6.
Spring rate
k=
3  E I
3
L
3.
6
E  30 10  psi
See Figure B-1(a) in Appendix B and Mathcad file P0439.
Spring rate
2.
Modulus of rigidity
Rod width
 4 L3 k 

a  
 E 
Volume of rod
V  a  L
2
I=
1
a
4
12
4
a  1.232  in
3
V  18.215 in
It takes slightly more material when a round rod is used than when a square rod is used. Thus, the square rod
is more efficient.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-40-1
PROBLEM 4-40
Statement:
Redesign the roll support of Problem 4-8 to be like that shown in Figure P4-16. The stub
mandrels insert to 10% of the roll length at each end. Choose appropriate dimensions a and
b to fully utilize the mandrel's strength, which is the same as in Problem 4-27. See Problem
4-8 for additional data.
Given:
Paper roll dimensions
Roll density
OD  1.50 m
S y  100  MPa
Material properties
ID  0.22 m
S ys  50 MPa
Lroll  3.23 m
E  207  GPa
3
ρ  984  kg m
Assumptions: 1. The paper roll's weight creates a concentrated load acting at the tip of the mandrel.
2. The mandrel's root in the stanchion experiences a distributed load over its length of
engagement
Solution:
See Figures 4-40 and Mathcad file P0440.
1. Model the support in such a way that stresses in the
portion of the mandrel that is inside the stanchion can
be determined. There are several assumptions that can
be made about the loads on this portion of the mandrel.
Figure 4-40A shows the one that will be used for this
design.
w
a
x
2. Determine the weight of the roll, the load on each
support, and the length of the mandrel.
W 
π
4

2
2

 OD  ID  Lroll  ρ  g
F
y
Lm
b
R
W  53.9 kN
FIGURE 4-40A
Free Body Diagram used in Problem 4-40
F  0.5 W
F  26.95  kN
Lm  0.1 Lroll
Lm  323  mm
3. From inspection of Figure 4-40A, write the load function equation
q(x) = -w<x>0 + w<x - b>0 + R<x - b>-1 - F<x - b -Lm>-1
4. Integrate this equation from - to x to obtain shear, V(x)
V(x) = -w<x>1 + w<x - b>1 + R<x - b>0 - F<x - b -Lm>0
5. Integrate this equation from - to x to obtain moment, M(x)
M(x) = -(w/2)<x>2 + (w/2)<x - b>2 + R<x - b>1 - F<x - b -Lm>1
6. Solve for the reactions by evaluating the shear and moment equations at a point just to the right of x = b + Lm,
where both are zero.
At x = (b + Lm)+ , V = M = 0
0 = w  b  Lm  w  Lm  R  F
R = F  w b
w
w
2 w
2
2 w
2
0 =    b  Lm   Lm  R Lm =    b  Lm   Lm  ( F  w b )  Lm
2
2
2
2
w=
2  F  Lm
b
2
Note that R is inversely proportional to b and w is inversly proportional to b 2.
7. To see the value of x at which the shear and moment are maximum, let
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
b  200  mm
w 
then
2  F  Lm
b
4-40-2
R  F  w b
and
2
L  b  Lm
x  0  mm 0.002  L  L
8. Define the range for x
9. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
S ( x z)  if ( x  z 1 0 )
10. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the
effect of the singularity functions.
V ( x)  w S ( x 0  mm)  x  w S ( x b )  ( x  b )  R S ( x b )  F  S ( x L)
M ( x) 
w
2
w
2
 S ( x 0  mm)  x 
2
2
 S ( x b )  ( x  b )  R S ( x b )  ( x  b )  F  S ( x L)  ( x  L)
11. Plot the shear and moment diagrams.
Shear Diagram
Moment Diagram
200
2
100
1
V ( x)
kN
M ( x)
0
kN  m
 100
 200
4
7
0
100
200
300
400
500
 10
600
0
100
200
300
x
x
mm
mm
400
500
600
FIGURE 4-40B
Shear and Moment Diagram Shapes for Problem 4-40
12. From Figure 4-40B, the maximum internal shear and moment occur at x = b and are
Vmax =
2  F  Lm
Mmax  F  Lm
b
Mmax  8.704  kN  m
13. The bending stress will be a maximum at the top or bottom of the mandrel at a section through x = b.
σmax =
Mmax a
2 I
where
I=
π a
64
4
so,
σmax =
π a
1
 32 W  Lm 
 π S 
y 

Solving for a,
a 
Round this to
a  125  mm
32 Mmax
3
= Sy
3
a  121.037  mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-40-3
14. Using this value of a and equation 4.15c, solve for the shear stress on the neutral axis at x = b.
τmax =
4  Vmax
3 A
=
8  F  Lm
 π a 2 
 b
3 
 4 
8  F  Lm
Solving for b
b 
Round this to
b  38 mm
= S ys
b  37.828 mm
 π a 2 
  Sys
3 
 4 
15. These are minimum values for a and b. Using them, check the bearing stress.
Magnitude of distributed load
w 
2  F  Lm
b
Bearing stress
σbear 
2
w b
a b
w  12055 
N
mm
σbear  96.4 MPa
Since this is less than S y, the design is acceptable for a  125  mm and
b  38 mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-41-1
PROBLEM 4-41
Statement:
A 10-mm ID steel tube carries liquid at 7 MPa. Determine the principal stresses in the wall if its
thickness is: a) 1 mm, b) 5 mm.
Given:
Tubing ID
Assumption:
The tubing is long therefore the axial stress is zero.
Solution:
See Mathcad file P0441.
(a) Wall thickness is
ID  10 mm
Inside pressure
p i  7  MPa
t  1  mm
1. Check wall thickness to radius ratio to see if this is a thick or thin wall problem.
ratio 
t
0.5 ID
ratio  0.2
Since the ratio is greater than 0.1, this is a thick wall problem.
2. Using equations 4.48a and 4.48b, the stresses are maximum at the inside wall where
Inside radius
ri  0.5 ID
ri  5  mm
Outside radius
ro  ri  t
ro  6  mm
Tangential stress
2

ro 


σt 
 1
2
2 
2
ri 
ro  ri 
σt  38.82  MPa
Radial stress
σr 
2
ri  p i

2
ri  p i
1 

ro  ri 
2
2
ro
ri
2

2
σr  7.00 MPa

3. Determine the principal stresses (since, for this choice of coordinates, the shear stress is zero),
σ1  σt
σ1  38.82  MPa
σ2  0  MPa
σ3  σr
σ3  7.00 MPa
The maximum shear stress is
τmax 
σ1  σ3
2
τmax  22.91  MPa
t  5  mm
(b) Wall thickness is
1. Check wall thickness to radius ratio to see if this is a thick or thin wall problem.
ratio 
t
0.5 ID
ratio  1
Since the ratio is greater than 0.1, this is a thick wall problem.
2. Using equations 4.48a and 4.48b, the stresses are maximum at the inside wall where
Inside radius
ri  0.5 ID
ri  5  mm
Outside radius
ro  ri  t
ro  10 mm
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-41-2
Tangential stress
2

ro 


σt 
 1
2
2 
2
ri 
ro  ri 
σt  11.67  MPa
Radial stress
2

ro 


σr 
 1
2
2 
2
ri 
ro  ri 
σr  7.00 MPa
2
ri  p i
2
ri  p i
3. Determine the principal stresses (since, for this choice of coordinates, the shear stress is zero),
σ1  σt
σ1  11.67  MPa
σ2  0  MPa
σ3  σr
σ3  7.00 MPa
The maximum shear stress is
τmax 
σ1  σ3
2
τmax  9.33 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-42-1
PROBLEM 4-42
Statement:
A cylindrical tank with hemispherical ends is required to hold 150 psi of pressurized air at room
temperature. Find the principal stresses in the 1-mm-thick wall if the tank diameter is 0.5 m and
its length is 1 m.
Given:
Tank ID
Wall thickness
Inside pressure
Solution:
1.
ID  500  mm
t  1  mm
p i  150  psi
p i  1034 kPa
See Mathcad file P0442.
Check wall thickness to radius ratio to see if this is a thick or thin wall problem.
ratio 
3
t
ratio  4  10
0.5 ID
Since the ratio is less than 0.1, this is a thin wall problem.
2.
3.
Using equations 4.49a, 4.49b and 4.49c, the stresses are
Radius
r  0.5 ID
Tangential stress
σt 
pi  r
t
Radial stress
σr  0  MPa
Axial stress
σa 
pi  r
2 t
r  250  mm
σt  258.55 MPa
σr  0.00 MPa
σa  129.28 MPa
Determine the principal stresses (since, for this choice of coordinates, the shear stress is zero),
σ1  σt
σ1  259  MPa
σ1  37.5 ksi
σ2  σa
σ2  129  MPa
σ2  18.75  ksi
σ3  0  MPa
σ3  0.00 MPa
σ3  0.00 MPa
The maximum shear stress is
τmax 
σ1  σ3
2
τmax  129  MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-43-1
PROBLEM 4-43
Statement:
Figure P4-17 shows an off-loading station at the end of a paper rolling machine. The finished
paper rolls are 0.9-m OD by 0.22-m ID by 3.23-m long and have a density of 984 kg/m3. The
rolls are transfered from the machine conveyor (not shown) to the forklift truck by the
V-linkage of the off-load station, which is rotated through 90 deg by an air cylinder. The paper
then rolls onto the waiting forks of the truck. The forks are 38-mm thick by 100-mm wide by
1.2-m long and are tipped at a 3-deg angle from the horizontal. Find the stresses in the two
forks on the truck when the paper rolls onto it under two different conditions (state all
assumptions):
(a) The two forks are unsupported at their free end.
(b) The two forks are contacting the table at point A.
Given:
Paper roll dimensions
OD  0.90 m
ID  0.22 m
Lroll  3.23 m
Roll density
ρ  984  kg m
Fork dimensions t  38 mm
w  100  mm
Lfork  1200 mm
3
θfork  3  deg
Assumptions: 1. The greatest bending moment will occur when the paper roll is at the tip of the fork for case (a)
and when it is midway between supports for case (b).
2. Each fork carries 1/2 the weight of a paper roll.
3. For case (a), each fork acts as a cantilever beam (see Appendix B-1(a)).
4. For case (b), each fork acts as a beam that is built-in at one end and simply-supported at the
other.
Solution:
See Figure 4-43 and Mathcad file P0443.
F
1. Determine the weight of the roll and the load on each
fork.
π
W 
4

2
2

 OD  ID  Lroll  ρ  g
F  0.5 W
L fork
t
W  18.64  kN
R1
F  9.32 kN
M1
Case (a), Cantilever Beam
2. The moment of inertia and the distance to the extreme
fiber for a fork are
I 
c 
w t
3
12
t
2
5
I  4.573  10  mm
4
0.5 L fork
t
c  19 mm
L fork
R1
Case (a)
R2
M2
Case (b), Fixed-Simply Supported Beam
3. From Figure D-1(a), the moment is a maximum at the
support and is
Mmax  F  Lfork
F
Mmax  11.186 kN  m
4. The bending stress is maximum at the support and is
FIGURE 4-43A
Free Body Diagrams used in Problem 4-43
σa 
Mmax c
I
σa  464.8  MPa
Case (b)
5. This beam is statically indeterminate. However, using singularity functions and the method shown in Example
4-7, we can determine the reactions and find the maximum moment.
6. Calculate the distance from the left support to the load and the distance between supports.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
a  0.5 Lfork
a  600  mm
L  Lfork
L  1200 mm
4-43-2
7. From inspection of Figure 4-43A, write the load function equation
q(x) = R1<x>-1 - F<x - a>-1 + R2<x - L>-1 + M2<x - L>-2
8. Integrate this equation from - to x to obtain shear, V(x)
V(x) = R1<x>0 - F<x - a>0 + R2<x - L>0 + M2<x - L>-1
9. Integrate this equation from - to x to obtain moment, M(x)
M(x) = R1<x>1 - F<x - a>1 + R2<x - L>1 + M2<x - L>0
10. Integrate the moment function, multiplying by 1/EI, to get the slope.
(x) = [R1<x>2/2 - F<x - a>2/2 + R2<x - L>2/2 + M2<x - L>1 + C3]/EI
11. Integrate again to get the deflection.
y(x) = [R1<x>3/6 - F<x - a>3/6 + R2<x - L>3/6 + M2<x - L>2/2 + C3x + C4]/EI
12. Evaluate R1, R2, M2, C3 and C4
At x = 0 and x = L; y = 0, therefore, C4 = 0. At x = L,  = 0
At x = L+, V = M = 0
Guess
R1  1  kN
R2  1  kN
M2  1  kN  m
2
C3  1  kN  m
Given
3
R 1 L
6

2
R 1 L
2

F  ( L  a)
3
6
F  ( L  a)
3
 C3 L = 0  kN  m
2
2
2
 C3 = 0  kN  m
R1  R2  F = 0  kN
R1 L  F  ( L  a )  M2 = 0  kN  m
 R1 
R 
 2   Find  R R M C 
1 2
2
3
 M2 
 
 C3 
R1  2.913  kN
13. Define the range for x
R2  6.409  kN
M2  2.097  kN  m
2
C3  0.419  kN  m
x  0  in 0.002  L  L
14. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than
z, and a value of one when it is greater than or equal to z.
S ( x z)  if ( x  z 1 0 )
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-43-3
15. Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the
effect of the singularity functions.
V ( x)  R1 S ( x 0  in)  F  S ( x a )  R2 S ( x L)
M ( x)  R1 S ( x 0  in)  x  F  S ( x a )  ( x  a )  R2 S ( x L)  ( x  L)
16. Plot the shear and moment diagrams.
Shear Diagram
Moment Diagram
10
2
1
5
V ( x)
kN
M ( x)
0
kN  m
1
5
 10
0
2
0
200
400
600
800
1000 1200
3
0
200
400
600
x
x
mm
mm
800
1000 1200
FIGURE 4-43B
Shear and Moment Diagrams for Problem 4-43
17. The maximum moment occurs at x = L,
Mmax  M ( L)
18. The bending stress is maximum at the support and is
Mmax  2.097  kN  m
σa 
Mmax c
I
σa  87.2 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-44-1
PROBLEM 4-44
Statement:
Determine a suitable thickness for the V-links of the off-loading station of Figure P4-17 to limit
their deflections at the tips to 10-mm in any position during their rotation. Two V-links support
the roll, at the 1/4 and 3/4 points along the roll's length, and each of the V arms is 10-cm wide by
1-m long. The V arms are welded to a steel tube that is rotated by the air cylinder. See Problem
4-43 for more information.
Given:
Roll OD
OD  0.90 m
Arm width
wa  100  mm
Roll ID
ID  0.22 m
Arm length
La  1000 mm
Roll length
Lroll  3.23 m
Max tip deflection
δtip  10 mm
Roll density
ρ  984  kg m
Mod of elasticity
E  207  GPa
3
Assumptions: 1. The maximum deflection on an arm will occur just after it begins the transfer and just before it
completes it, i.e., when the angle is either zero or 90 deg., but after the tip is no longer
supported by the base unit.
2. At that time the roll is in contact with both arms ("seated" in the V) and will remain in that
state throughout the motion. When the roll is in any other position on an arm the tip will be
supported.
3. The arm can be treated as a cantilever beam with nonend load.
4. A single arm will never carry more than half the weight of a roll.
5. The pipe to which the arms are attached has OD = 160 mm.
Solution:
See Figure 4-44 and Mathcad file P0444.
1. Determine the weight of the roll and the load on each
V-arm.
W 

4
π
2
2

 OD  ID  Lroll  ρ  g
450
W  18.64  kN
F  0.5 W
F  9.32 kN
2. From Appendix B, Figure B-1, the tip deflection of a
cantilever beam with a concentrated load located at a
distance a from the support is
ymax =
F a
2
6  E I
 ( a  3  L)
1000 = L
370 = a
where L is the beam length and I is the
cross-section moment of inertia. In this case
M
3
I=
3. Setting
w a t a
F
12
ymax = δtip
F
FIGURE 4-44
a  370  mm
and
Free Body Diagram used in Problem 4-44
substituting for I and solving for ta
1
 2 F  a2  3 La  a 
ta 


E δtip  wa


Let the arm thickness be
3
ta  31.889 mm
ta  32 mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-45-1
PROBLEM 4-45
Statement:
Determine the critical load on the air cylinder rod in Figure P4-17 if the crank arm that rotates it is
0.3 m long and the rod has a maximum extension of 0.5 m. The 25-mm-dia rod is solid steel with a
yield strength of 400 MPa. State all assumptions.
Given:
Rod length
Rod diameter
L  500  mm
d  25 mm
E  207  GPa
S y  400  MPa
Young's modulus
Yield strength
Assumptions: 1. The rod is a fixed-pinned column.
2. Use a conservative value of 1 for the end factor (see Table 4-7 in text).
Solution:
1.
See Mathcad file P0445.
Calculate the slenderness ratio that divides the unit load vs slenderness ratio graph into Johnson and Euler
regions.
S rD  π
2.
2 E
Calculate the cross-section area and the moment of inertia.
Area
Moment of inertia
3.
S rD  101.07
Sy
A 
π 2
I 
π
d
4
64
d
4
4
2
I  1.92  10  mm
4
Using Table 4-7, calculate the effective column length.
Leff  1  L
4.
A  490.87 mm
Leff  500  mm
Calculate the slenderness ratio for the column.
Radius of gyration
Slenderness ratio
k 
S r 
I
A
Leff
k
k  6.25 mm
S r  80.00
Since the Sr for this column is less than SrD, it is a Johnson column.
5.
Calculate the critical load using the Johnson equation.
2

1  S y S r  

Pcr  A  S y   

E  2 π  

Pcr  134.8  kN
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-46-1
PROBLEM 4-46
Statement:
The V-links of Figure P4-17 are rotated by the crank arm through a shaft that is 60 mm dia by
3.23 m long. Determine the maximum torque applied to this shaft during motion of the
V-linkage and find the maximum stress and deflection for the shaft. See Problem 4-43 for more
information.
Given:
Paper roll dimensions
OD  900  mm
ID  220  mm
Shaft dims
d  60 mm
Lshaft  3230 mm
Lroll  3230 mm
3
ρ  984  kg m
Roll density
G  79 GPa
Modulus of rigidity
Assumptions: The greatest torque will occur when the link is horizontal and the paper roll is located as shown
in Figure P4-17 or Figure 4-46.
Solution:
See Figure 4-46 and Mathcad file P0446.
y
1. Determine the weight of the roll on the V-arms.

4
π
W 
2
2

 OD  ID  Lroll  ρ  g
W  18.64  kN
2. Summing moments about the shaft center,
T 
OD
2
W
T  8.390  kN  m
3. Calculate the polar moment of inertia.
J 
π d
4
W
6
J  1.272  10  mm
32
4
T
Ry
4. The maximum torsional stress will be at the outside
diameter of the shaft. The radius of the OD is,
r 
d
r  30 mm
2
60-mm-dia shaft
450.0
FIGURE 4-46
Free Body Diagram used in Problem 4-46
5. Determine the maximum torsional stress using equation (4.23b).
τmax 
Tr
J
τmax  197.8  MPa
6. Use equation (4.24) to determine the angular shaft deflection.
θ 
T  Lshaft
J G
θ  15.447 deg
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-47-1
PROBLEM 4-47
Statement:
Determine the maximum forces on the pins at each end of the air cylinder of Figure P4-17.
Determine the stress in these pins if they are 30-mm dia and in single shear.
Given:
Paper roll dimensions
OD  0.90 m
ID  0.22 m
Lroll  3.23 m
Roll density
ρ  984  kg m
Pin diameter
d  30 mm
3
Assumptions: 1. The maximum force in the cylinder rod occurs when the transfer starts.
2. The cylinder and rod make an angle of 8 deg to the horizontal at the start of transfer.
3. The crank arm is 300 mm long and is 45 deg from vertical at the start of transfer.
4. The cylinder rod is fully retracted at the start of the transfer. At the end of the transfer it will
have extended 500 mm from its initial position.
Solution:
See Figure 4-47 and Mathcad file P0447.
1. Determine the weight of the roll on the
forks.
W 

4
π
2
2
y

 OD  ID  Lroll  ρ  g
W  18.64  kN
2. From the assumptions and
Figure 4-47, the x and y
distances from the origin to
point A are,
Rax  300  cos( 45 deg)  mm
W
Ray  300  sin( 45 deg)  mm
Rx
Rax  212.132  mm
x
212.1
Ry
A
Ray  212.132  mm
F
8°
212.1
450.0
3. From Figure 4-47, the x
distance from the origin to
point where W is applied is,
FIGURE 4-47
Free Body Diagram at Start of Transfer for V-link of Problem 4-47
Rwx 
4.
OD
2
Rwx  450  mm
Sum moments about the pivot point and solve for the compressive force in the cylinder rod.
W  Rwx  F  Rax sin( 8  deg)  F  Ray cos( 8  deg) = 0
F 
W  Rwx
Ray cos( 8  deg)  Rax sin( 8  deg)
F  46.469 kN
This is the shear force in the pins
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5.
4-47-2
Determine the cross-sectional area of the pins and the direct shear stress.
Shear area
Shear stress
A 
τ 
π d
4
F
A
2
A  706.858  mm
2
τ  65.7 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-48-1
PROBLEM 4-48
Statement:
A 100-kg wheelchair marathon racer wants an exerciser that will allow indoor practicing in any
weather. The design shown in Figure P4-18 is proposed. Two free-turning rollers on bearings
support the rear wheels. A platform supports the front wheels. Design the 1-m-long rollers as
hollow tubes of aluminum to minimize the height of the platform and also limit the roller
deflections to 1 mm in the worst case. The wheelchair has 650-mm-dia drive wheels separated
by a 700-mm track width. The flanges shown on the rollers limit the lateral movement of the
chair while exercising and thus the wheels can be anywhere between those flanges. Specify
suitable sized steel axles to support the tubes on bearings. Calculate all significant stresses.
Given:
Mass of chair
M  100  kg
Wheel diameter d w  650  mm
Track width
T  700  mm
Aluminum
Ea  71.7 GPa
Roller length
Lr  1000 mm
Steel
Es  207  GPa
Assumptions: 1. The CG of the chair with rider is
sufficiently close to the rear wheel that all of
the weight is taken by the two rear wheels.
2. The small camber angle of the rear wheels
does not significantly affect the magnitude
of the forces on the rollers.
3. Both the aluminum roller and the steel axle
are simply supported. The steel axles that
support the aluminum tube are fixed in the
mounting block and do not rotate. The
aluminum tube is attached to them by two
bearings (one on each end of the tubes, one
for each axle). The bearings' inner race is
fixed, and the outer race rotates with the
aluminum tube. Each steel axle is considered
to be loaded as a simply supported beam.
Their diameter must be less than the inner
diameter of the tubes to fit the roller bearings
between them.
Solution:
δ  1  mm
Maximum deflection
Modulus elasticity
F


Free Body Diagram of One Wheel
used in Problem 4-48
1. Calculate the weight of the chair with rider.
W  M  g
F
FIGURE 4-48A
See Figures 4-48 and Mathcad file P0448.
Weight of chair
W/2
W  980.7  N
2. Calculate the forces exerted by the wheels on the rollers (see Figure 4-48A). From the FBD of a wheel,
summing vertical forces
2  F  cos( θ ) 
Let
3.
θ  20 deg
W
2
=0
then
F 
W
4  cos( θ )
F  260.9  N
The worst condition (highest moment at site of a stress concentration) will occur when the chair is all the
way to the left or right. Looking at the plane through the roller that includes the forces exerted by the wheels
(the FBD is shown in Figure 4-48B) the reactions R1 and R2 come from the bearings, which are inside the
hollow roller and are, themselves, supported by the steel axle.
4. Solving for the reactions. Let the distance from R1 to F be a  15 mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
 M1
R2 Lr  F  ( a  T )  F  a = 0
 Fy
R1  2  F  R2 = 0
R2 
F  (2 a  T )
4-48-2
700
F
F
R2  190.5  N
Lr
15
R1  2  F  R2
R2
R1
R1  331.3  N
1000
FIGURE 4-48B
Free Body Diagram of One Tube used in Problem 4-48
5. The maximum bending moment will be at the right-hand load and will be
Mrmax  R2 Lr  ( a  T )
Mrmax  54.3 N  m
Note, if the chair were centered on the roller the maximum moment would be
Mc  F 
Lr  T
Mc  39.1 N  m
2
and this would be constant along the axle between the two loads, F.
6. Note that the bearing positions are fixed regardless of the position of the chair on the roller.
Because of symmetry,
Ra1  R1
Ra1  331.3  N
Ra2  R2
Ra2  190.5  N
1000
65
R1
7. The maximum bending moment
occurs
at R1 and is for b  65 mm
Mamax  Ra1 b
R2
R a1
R a2
1130
FIGURE 4-48C
Free Body Diagram of One Axle used in Problem 4-48
Mamax  21.5 N  m
8. Determine a suitable axle diameter. Let the factor of safety against yielding in the axle be
Nsa  3
9. Tentatively choose a low-carbon steel for the axle, say AISI 1020, cold rolled with
S y  393  MPa
10. At the top of the axle under the load R1 there is only a bending stress. Set this stress equal to the yield
strength divided by the factor of safety.
σx =
32 Mamax
3
π d a
=
Sy
Nsa
1
 32 Nsa Mamax 


π S y


Solving for the axle diameter, d a
d a 
Let the axle diameter be
d a  15 mm
3
d a  11.875 mm
made from cold-rolled AISI 1020 steel.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
11. Suppose that bearing 6302 from Chapter
10, Figure 10-23. It has a bore of 15 mm and an
OD of 42 mm. Thus, the inside diameter of the
roller away from the bearings where the
moment is a maximum will be d i  40 mm.
This will provide a 1-mm shoulder for axial
location of the bearings.
4-48-3
150
700
F
F
F
15
12. The maximum deflection of the roller will
occur when the chair is in the center of the
roller. For this case the reactions are both equal
to the loads, F (see Figure 4-48D). The
maximum deflection is at the center of the roller.
F
1000
FIGURE 4-48D
Free Body Diagram of Roller with Chair in the Center.
13. Write the load function and then integrate four times to get the deflection function.
q(x) = F<x>-1 - F<x - a>-1 - F<x - b>-1 + F<x - L>-1
y(x) = F[<x>3 - <x - a>3 - <x - b>3 + <x - L>3 + C3x]/(6EI)
where
C3 =
1
L
 ( L  a )  a  L 
3
3
3
14. Write the deflection function at x = L/2 for
a  150  mm
ymax
 L  3
=
   
6  Ea I  2 
F
3
 L  a  1  ( L  a) 3  a3  L3
2

2



15. Set this equation equal to the allowed deflection  and solve for the required moment of inertia, I.
 Lr  3
I 
   
6  Ea δ  2 
F
3

 Lr

1 
3
3
3
  a    Lr  a   a  Lr 
2
2


4
I  6.618  10  mm
4
16. Knowing the inside diameter of the tube, solve for the outside diameter.
1
π  4
4
I=
  d o  d i 
64
Round this up to
64 I
4
d o  
 d i 
 π

4
d o  44.463 mm
d o  46 mm
DESIGN SUMMARY
Axles
Rollers
Material
AISI 1020 steel, cold-rolled
Material
2024-T4 aluminum
Diameter
d a  15 mm
Outside diameter
d o  46 mm
Length
1220 mm
Inside diameter
d i  40 mm
Length
1040 mm
Spacing
c   d w  d o  sin( θ )
c  238  mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-49a-1
PROBLEM 4-49a
Statement:
A hollow, square column has the dimensions and properties below. Determine if it is a Johnson
or an Euler column and find the critical load:
(a) If its boundary conditions are pinned-pinned.
(b) If its boundary conditions are fixed-pinned.
(c) If its boundary conditions are fixed-fixed.
(d) If its boundary conditions are fixed-free.
Given:
Length of column
Outside dimension
Material
L  100  mm
Yield strength
so  4  mm
Steel
S y  300  MPa
Inside dimension
si  3  mm
E  207  GPa
Solution:
1.
See Mathcad file P0449a.
Calculate the slenderness ratio that divides the unit load vs slenderness ratio graph into Johnson and Euler
regions.
S rD  π
2.
Modulus of elasticity
2 E
S rD  116.7
Sy
Calculate the cross-section area and the moment of inertia.
2
Area
A  so  si
Moment of inertia
I 
1
12

2
4
 so  si
A  7.00 mm

4
2
I  14.58  mm
4
(a) pinned-pinned ends
3.
Using Table 4-7, calculate the effective column length.
Leff  1  L
4.
Leff  100  mm
Calculate the slenderness ratio for the column.
Radius of gyration
k 
S r 
Slenderness ratio
I
k  1.443  mm
A
Leff
k
S r  69.28
Since the S r for this column is less than S rD, it is a Johnson column.
5.
Calculate the critical load using the Johnson equation.

Pcr  A  S y 

2
 Sy S r  

E  2 π  
1

Pcr  1.73 kN
(b) fixed-pinned ends
6.
Using Table 4-7, calculate the effective column length.
Leff  0.8 L
7.
Leff  80 mm
Calculate the slenderness ratio for the column.
Radius of gyration
k 
I
A
k  1.443  mm
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
Slenderness ratio
S r 
Leff
k
4-49a-2
S r  55.43
Since the S r for this column is less than S rD, it is a Johnson column.
8.
Calculate the critical load using the Johnson equation.
2

1  S y S r  

Pcr  A  S y   

E  2 π  

Pcr  1.86 kN
(c) fixed-fixed ends
9.
Using Table 4-7, calculate the effective column length.
Leff  0.65 L
Leff  65 mm
10. Calculate the slenderness ratio for the column.
Radius of gyration
Slenderness ratio
k 
S r 
I
A
Leff
k
k  1.443  mm
S r  45.03
Since the S r for this column is less than S rD, it is a Johnson column.
11. Calculate the critical load using the Johnson equation.
2

S y S r  
1

Pcr  A  S y   

E  2 π  

Pcr  1.94 kN
(d) fixed-free ends
12. Using Table 4-7, calculate the effective column length.
Leff  2.1 L
Leff  210  mm
13. Calculate the slenderness ratio for the column.
Radius of gyration
Slenderness ratio
k 
S r 
I
A
Leff
k
k  1.443  mm
S r  145.49
Since the S r for this column is greater than S rD, it is an Euler column.
14. Calculate the critical load using the Euler equation.
2
Pcr  A 
π E
Sr
2
Pcr  676  N
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-50a-1
PROBLEM 4-50a
Statement:
A hollow, round column has the dimensions and properties below. Determine if it is a Johnson
or an Euler column and find the critical load:
(a) If its boundary conditions are pinned-pinned.
(b) If its boundary conditions are fixed-pinned.
(c) If its boundary conditions are fixed-fixed.
(d) If its boundary conditions are fixed-free.
Given:
Length of column
Outside diameter
L  1500 mm Material
Yield strength
od  20 mm
Steel
S y  300  MPa
Inside diameter
id  14 mm
E  207  GPa
Solution:
1.
See Mathcad file P0450a.
Calculate the slenderness ratio that divides the unit load vs slenderness ratio graph into Johnson and Euler
regions.
S rD  π
2.
3.
Modulus of elasticity
2 E
S rD  116.7
Sy
Calculate the cross-section area, moment of inertia, and the radius of gyration.
Area
A 
Moment of inertia
I 
Radius of gyration
k 

4
2
2

64
4
4
π
π
 od  id
 od  id

A  160.22 mm

I  5968 mm
I
2
4
k  6.103  mm
A
Define functions to determine column type and critical load.
Type
type S r 
"Euler" if S r  S rD
"Johnson" otherwise
Critical load
Pcr S r 
2
return A 
π E
Sr
2
if type S r = "Euler"
2

1  S y S r  

A  Sy   
 otherwise
E  2 π  

(a) pinned-pinned ends
4.
Using Table 4-7, calculate the effective column length.
Leff  1  L
5.
Calculate the slenderness ratio for the column.
Slenderness ratio
6.
Leff  1500 mm
S r 
Leff
k
S r  245.77
Determine the type and critical load using the functions defined above.
type S r  "Euler"
Pcr S r  5.42 kN
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-50a-2
(b) fixed-pinned ends
7.
Using Table 4-7, calculate the effective column length.
Leff  0.8 L
8.
Calculate the slenderness ratio for the column.
Slenderness ratio
9.
Leff  1200 mm
S r 
Leff
k
S r  196.62
Determine the type and critical load using the functions defined above.
type S r  "Euler"
Pcr S r  8.47 kN
(c) fixed-fixed ends
10. Using Table 4-7, calculate the effective column length.
Leff  0.65 L
Leff  975  mm
11. Calculate the slenderness ratio for the column.
Slenderness ratio
S r 
Leff
k
S r  159.75
12. Determine the type and critical load using the functions defined above.
type S r  "Euler"
Pcr S r  12.8 kN
(d) fixed-free ends
13. Using Table 4-7, calculate the effective column length.
Leff  2.1 L
Leff  3150 mm
14. Calculate the slenderness ratio for the column.
Slenderness ratio
S r 
Leff
k
S r  516.12
15. Determine the type and critical load using the functions defined above.
type S r  "Euler"
Pcr S r  1.23 kN
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-51a-1
PROBLEM 4-51a
Statement:
A solid, rectangular column has the dimensions and properties below. Determine if it is a
Johnson or an Euler column and find the critical load:
(a) If its boundary conditions are pinned-pinned.
(b) If its boundary conditions are fixed-pinned.
(c) If its boundary conditions are fixed-fixed.
(d) If its boundary conditions are fixed-free.
Given:
Length of col.
Thickness
L  100  mm
t  10 mm
Material
Yield strength
Steel
S y  300  MPa
Height
h  20 mm
Modulus of elasticity
E  207  GPa
Solution:
1.
See Mathcad file P0451a.
Calculate the slenderness ratio that divides the unit load vs slenderness ratio graph into Johnson and Euler
regions.
S rD  π
2.
2 E
Calculate the cross-section area, moment of inertia, and the radius of gyration.
Area
A  h  t
Moment of inertia
I 
Radius of gyration
3.
S rD  116.7
Sy
k 
h t
A  200.00 mm
3
I  1667 mm
12
I
2
4
k  2.887  mm
A
Define functions to determine column type and critical load.
Type
type S r 
"Euler" if S r  S rD
"Johnson" otherwise
Critical load
Pcr S r 
2
return A 
π E
Sr

A  S y 

2
if type S r = "Euler"
2
 S y Sr  
 otherwise
E  2 π  
1

(a) pinned-pinned ends
4.
Using Table 4-7, calculate the effective column length.
Leff  1  L
5.
Calculate the slenderness ratio for the column.
Slenderness ratio
6.
Leff  100  mm
S r 
Leff
k
S r  34.64
Determine the type and critical load using the functions defined above.
type S r  "Johnson"
Pcr S r  57.36  kN
© 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,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-51a-2
(b) fixed-pinned ends
7.
Using Table 4-7, calculate the effective column length.
Leff  0.8 L
8.
Calculate the slenderness ratio for the column.
Slenderness ratio
9.
Leff  80 mm
S r 
Leff
k
S r  27.71
Determine the type and critical load using the functions defined above.
type S r  "Johnson"
Pcr S r  58.31  kN
(c) fixed-fixed ends
10. Using Table 4-7, calculate the effective column length.
Leff  0.65 L
Leff  65 mm
11. Calculate the slenderness ratio for the column.
Slenderness ratio
S r 
Leff
k
S r  22.52
12. Determine the type and critical load using the functions defined above.
type S r  "Johnson"
Pcr S r  58.9 kN
(d) fixed-free ends
13. Using Table 4-7, calculate the effective column length.
Leff  2.1 L
Leff  210  mm
13. Calculate the slenderness ratio for the column.
Slenderness ratio
S r 
Leff
k
S r  72.75
14. Determine the type and critical load using the functions defined above.
type S r  "Johnson"
Pcr S r  48.34  kN
© 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.
MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-52a-1
PROBLEM 4-52a
Statement:
A solid, circular column, loaded eccentrically, has the dimensions and properties below. Find
the critical load:
(a) If its boundary conditions are pinned-pinned.
(b) If its boundary conditions are fixed-pinned.
(c) If its boundary conditions are fixed-fixed.
(d) If its boundary conditions are fixed-free.
Given:
Length of column
Outside diameter
L  100  mm
od  20 mm
Material
Yield strength
Steel
S y  300  MPa
Eccentricity (t)
e  10 mm
Modulus of elasticity
E  207  GPa
Solution:
1.
See Mathcad file P0452a.
Calculate the cross-section area, distance to extreme fiber, and the moment of inertia.
A 
Area
4.
π
4
2
 od
Distance to extreme fiber
c  0.5 od
Moment of inertia
I 
π
4
64
 od
A  314.16 mm
2
c  10 mm
I  7854 mm
4
Calculate the radius of gyration and eccentricity ratio for the column.
k 
Radius of gyration
Er 
Eccentricity ratio
I
A
e c
k
2
k  5.00 mm
Er  4.0
(a) pinned-pinned ends
3.
Using Table 4-7, calculate the effective column length.
Leff  1  L
4.
Leff  100  mm
Calculate the slenderness ratio for the column.
S r 
Slenderness ratio
5.
Leff
k
S r  20.00
Calculate the critical load using the Secant equation.
Guess
P  1  kN
Given
S y A
P=




4  E A 
P
1  Er sec S r
Pcr  Find ( P)
Pcr  18.63  kN
(b) fixed-pinned ends
6.
Using Table 4-7, calculate the effective column length.
Leff  0.8 L
7.
Leff  80 mm
Calculate the slenderness ratio for the column.
© 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,
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Pearson Education, Inc., Upper Saddle River, NJ 07458.
MACHINE DESIGN - An Integrated Approach, 4th Ed.
S r 
Slenderness ratio
8.
Leff
k
4-52a-2
S r  16.00
Calculate the critical load using the Secant equation.
Guess
P  1  kN
Given
S y A
P=




P
1  Er sec S r
4  E A 
Pcr  Find ( P)
Pcr  18.71  kN
(c) fixed-fixed ends
9.
Using Table 4-7, calculate the effective column length.
Leff  0.65 L
Leff  65 mm
10. Calculate the slenderness ratio for the column.
S r 
Slenderness ratio
Leff
k
S r  13.00
11. Calculate the critical load using the Secant equation.
Guess
P  1  kN
Given
S y A
P=


4  E A 


P
1  Er sec S r
Pcr  Find ( P)
Pcr  18.76  kN
(d) fixed-free ends
12. Using Table 4-7, calculate the effective column length.
Leff  2.1 L
Leff  210  mm
13. Calculate the slenderness ratio for the column.
S r 
Slenderness ratio
Leff
k
S r  42
14. Calculate the critical load using the Secant equation.
Guess
P  1  kN
Given
S y A
P=


1  Er sec S r
Pcr  Find ( P)


4  E A 
P
Pcr  17.93  kN
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-53-1
PROBLEM 4-53
Statement:
Design an aluminum, hollow, circular column for the conditions given below for (a)
pinned-pinned ends and (b) fixed-free ends.
Given:
Length of column
Wall thickness
Factor of safety
L  3  m
Yield strength
t  5  mm
FS  3
S yc  150  MPa
Load supported
F  900  N
E  71.7 GPa
Solution:
1.
See Mathcad file P0453.
Start by calculating the slenderness ratio that divides the unit load vs slenderness ratio graph into Johnson
and Euler regions.
2 E
S rD  π
2.
Modulus of elasticity
S rD  97.136
S yc
Using Table 4-7, calculate the effective column length.
Leff  1  L
3.
Leff  3000 mm
To start the iterative process, assume that the final design will be an Euler column with the critical load equal
to FS*F. From equation 4.38b,
2
Pcr =
π  E A  k
2
I
2
and
k =
2
A
L
2
Substituting for k2
Pcr =
π  E I
2
= FS  F
L
2
I 
Solving for I
Leff  FS  F
2
π E
The required moment of inertia, assuming an Euler column is I  34339  mm
4.
Using the relationships given on the inside cover, solve for the outside diameter of the tube.
Guess
D  20 mm
Given
I=
π  4
4
 D  ( D  2  t) 
64
D  Find ( D)
5.
4
D  30.64  mm
Using this diameter, calculate the slenderness ratio and compare to S rD. If it is greater than S rD the assumption o
an Euler column is correct, if not, recalculate using the Johnson equation.
Inside diameter
d  D  2  t
Area
Ar 
Radius of gyration
kr 
Slenderness ratio
S r 

4
π
2
 D d
I
Ar
Leff
kr
d  20.64  mm

2
Ar  402.7  mm
2
kr  9.234  mm
S r  324.9
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-53-2
Since this is greater than S rD, the assumption of an Euler column is correct and the minimum outside diameter is
D  30.64  mm
(b) fixed-free ends
6.
Using Table 4-7, calculate the effective column length.
Leff  2.1 L
7.
Leff  6300 mm
To start the iterative process, assume that the final design will be an Euler column with the critical load equal
to FS*F. From equation 4.38b,
2
Pcr =
π  E A  k
2
I
2
and
k =
2
A
L
2
Substituting for k2
Pcr =
π  E I
2
= FS  F
L
2
I 
Solving for I
Leff  FS  F
2
π E
5
The required moment of inertia, assuming an Euler column is I  2  10  mm
8.
Using the relationships given on the inside cover, solve for the outside diameter of the tube.
Guess
D  20 mm
Given
I=
π  4
4
 D  ( D  2  t) 
64
D  Find ( D)
9.
4
D  47.37  mm
Using this diameter, calculate the slenderness ratio and compare to S rD. If it is greater than S rD the
assumption of an Euler column is correct, if not, recalculate using the Johnson equation.
Inside diameter
d  D  2  t
Area
Ar 
Radius of gyration
kr 
Slenderness ratio
S r 

4
π
2
 D d
I
Ar
Leff
kr
d  37.37  mm

2
Ar  665.6  mm
2
kr  15.084 mm
S r  417.7
Since this is greater than S rD, the assumption of an Euler column is correct and the minimum outside diameter
is
D  47.37  mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-54-1
PROBLEM 4-54
Statement:
Three round, 1.25-in-dia bars are made of SAE 1030 hot-rolled steel but are of different lengths,
5 in, 30 in, and 60 in, respectively. They are loaded axially in compression. Compare the load
supporting capability of the three bars if the ends are assumed to be:
(a) Pinned-pinned.
(b) Fixed-pinned.
(c) Fixed-fixed.
(d) Fixed-free.
Given:
Outside diameter
Lengths
Material
Solution:
1.
2
SAE 1030 Steel
L  60 in
i  1 2  3
3
Yield strength
S y  38 ksi
Modulus of elasticity
E  30 10  psi
6
Calculate the slenderness ratio that divides the unit load vs. slenderness ratio graph into Johnson and Euler
regions.
2 E
S rD  124.8
Sy
Calculate the cross-section area, moment of inertia, and the radius of gyration.
Area
Moment of inertia
Radius of gyration
3.
L  30 in
1
See Mathcad file P0454.
S rD  π
2.
d  1.25 in
L  5  in
A 
π 2
I 
π
k 
d
4
64
d
A  791.73 mm
4
I  49882  mm
I
2
4
k  7.938  mm
A
Define functions to determine column type and critical load.
Type
type S r 
"Euler" if S r  S rD
"Johnson" otherwise
Critical load
Pcr S r 
2
return A 
π E
Sr
2
if type S r = "Euler"
2

S y S r  
1

A  S y   
  otherwise
E  2 π  

(a) pinned-pinned ends
4.
Using Table 4-7, calculate the effective column length.
Leff  1  L
5.
5
Leff   30   in
 
 60 
Calculate the slenderness ratio for the column.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
S r 
Slenderness ratio
6.
4-54-2
 16 
S r   96 
 
 192 
Leff
k
Determine the type and critical load using the functions defined above.
 "Johnson" 
Type   "Johnson" 
i 

 "Euler" 
 
Type  type S r
i
i
 
Pcr S r
i
lbf
46250
32844
9857
(b) fixed-pinned ends
7.
Using Table 4-7, calculate the effective column length.
 4.0 
Leff   24.0   in


 48.0 
Leff  0.8 L
8.
Calculate the slenderness ratio for the column.
Slenderness ratio
9.
S r 
 12.8 
S r   76.8 


 153.6 
Leff
k
Determine the type and critical load using the functions defined above.
 
Type  type S r
i
i
 "Johnson" 
Type   "Johnson" 
i 

 "Euler" 
 
Pcr S r
i
lbf
46388
37808
15401
(c) fixed-fixed ends
10.
Using Table 4-7, calculate the effective column length.
Leff  0.65 L
11.
 3.3 
Leff   19.5   in


 39.0 
Calculate the slenderness ratio for the column.
Slenderness ratio
S r 
Leff
k
 10.4 
S r   62.4 


 124.8 
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
12.
4-54-3
Determine the type and critical load using the functions defined above.
 
Type  type S r
i
i
 "Johnson" 
Type   "Johnson" 
i 

 "Johnson" 
 
Pcr S r
i
lbf
46471
40807
23329
(d) fixed-free ends
13.
Using Table 4-7, calculate the effective column length.
 10.5 
Leff   63.0   in


 126.0 
Leff  2.1 L
14.
Calculate the slenderness ratio for the column.
Slenderness ratio
15.
S r 
Leff
k
 33.6 
S r   201.6 


 403.2 
Determine the type and critical load using the functions defined above.
 
Type  type S r
i
i
 "Johnson" 
Type   "Euler" 
i 

 "Euler" 
 
Pcr S r
i
lbf
44944
8940
2235
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-55-1
PROBLEM 4-55
_____
Statement:
Figure P4-19 shows a 1.5-in-dia, 30-in-long steel rod subjected to tensile loads P = 10000 lb
applied at each end of the rod, acting along its longitudinal Y axis and through the centroid of
its circular cross section. Point A is 12 in below the upper end and point B is 8 in below A. For
this bar with its loading, find:
(a) All components of the stress tensor matrix (equation 4.1a) for a point midway between
A and B.
(b) The displacement of point B relative to point A.
(c) The elastic strain in the section between A and B.
(d) The total strain in the section between A and B.
Given:
Tensile load
P  10000  lbf
Diameter
Lengths
Modulus of elasticity
d  1.50 in
L  30 in
LA  12 in
LAB  8  in
Solution:
1.
See Mathcad file P0455.
Calculate the cross-section area of the rod.
A 
2.
6
E  30 10  psi
π d
2
2
A  1.767  in
4
(a) The loading is simple axial tension so all components of the stress tensor are zero except yy, which is
found using equation 4.7.
σyy 
P
σyy  5659 psi
A
This stress is uniform across the rod and has the same value at any cross section along the longitudinal axis
except close to the ends where the load is applied.
3.
(b) The displacement of point B relative to A can be found using equation 4.8.
ΔsBA 
4.
A E
ΔsBA  1.509  10
3
 in
(c) The elastic strain in the rod can be found using Hooke's law (equation 2.2)
ε 
5.
P LAB
σyy
E
ε  1.886  10
4
(d) Assuming that the yield strength of this steel is greater than yy, the strain calculated in step 4 is the total
strain.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-56-1
PROBLEM 4-56
_____
Statement:
The rod in Figure P4-19, with the loading of Problem 4-55, is subjected to a reduction of
temperature from 80F to 20F after the load is applied. The coefficient of thermal expansion for
steel is approximately 6 in/in/degF. Find:
(a) All components of the stress tensor matrix (equation 4.1a) for a point midway between
A and B.
(b) The displacement of point B relative to point A.
(c) The elastic strain in the section between A and B.
(d) The total strain in the section between A and B.
Units:
Temperature scale F  1
Given:
Tensile load
P  10000  lbf
Diameter
Lengths
Modulus of elasticity
d  1.50 in
L  30 in
LA  12 in
LAB  8  in
Temperatures
T1  80 F
Solution:
1.
Coefficient of thermal expansion
See Mathcad file P0456.
T2  20 F
α  6  10
6
F
1
Calculate the cross-section area of the rod.
A 
2.
6
E  30 10  psi
π d
2
2
A  1.767  in
4
(a) The loading is simple axial tension so all components of the stress tensor are zero except yy, which is
found using equation 4.7.
σyy 
P
σyy  5659 psi
A
This stress is uniform across the rod and has the same value at any cross section along the longitudinal axis
except close to the ends where the load is applied. The change in temperature does not affect the stress
since the ends are free.
3.
(b) The displacement of point B relative to A can be found by summing equation 4.8 for the elastic portion
and the thermal expansion equation from elementary mechanics of materials for the thermal portion.
ΔsBA 
4.
A E
 α  T2  T1  LAB
3
ΔsBA  1.371  10
 in
(c) The elastic strain in the rod can be found using Hooke's law (equation 2.2)
ε 
5.
P LAB
σyy
E
ε  1.886  10
4
(d) Assuming that there is no plastic strain in the rod, the total strain is the sum of the elastic strain found
in step 4 plus the thermal strain.
ε tot  ε  α  T2  T1
ε tot  1.714  10
4
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-57-1
PROBLEM 4-57
_____
Statement:
Figure P4-20 shows a steel bar fastened to a rigid ground plane with two 0.25-in-dia hardened
steel dowel pins. For P = 1500 lb, find:
(a) The shear stress in each pin.
(b) The direct bearing stress in each pin and hole.
(c) The minimum value of dimension h to prevent tearout failure if the steel bar has a
shear strength of 32500 psi.
Given:
Pin diameter
Distance between pins
d  0.250  in
a  2.0 in
Thickness of bar
t  0.25 in
Solution:
1.
Applied load
Shear strength of bar
Distance from right pin to load
P  1500 lbf
S s  32.5 ksi
b  4.0 in
See Mathcad file P0457.
Draw a free-body diagram and find the shear forces (reactions) on each pin.
a
b
RL
h
RR
P
Write equations 3.3b for the bar and solve for the reactions.
 F:
RL 
2.
b
a
P
π d
2
RR  P  RL
RR  4500 lbf
4
2
A  0.0491 in
(a) Use equation 4.9 to determine the shear stress in each pin.
Left pin
Right pin
4.
RL  3000 lbf
RL  a  P b  0
Calculate the cross-section area of a pin.
A 
3.
 M:
RL  RR  P  0
τL 
τR 
RL
τL  61.1 ksi
A
RR
τR  91.7 ksi
A
(b) Calculate the bearing area from equation 4.10 and use it to determine the bearing stress in each pin.
Bearing area
Abear  d  t
σL 
RL
Abear
2
Abear  0.0625 in
σL  48.0 ksi
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
σR 
5.
RR
Abear
4-57-2
σR  72.0 ksi
h  d

  t , where (h - d)/2 is the distance from the edge of the hole to
2

 
the outside of the bar. Substitute this area in equation 4.9 for the shear area and substitute the shear
strength for xy, solving then for the unknown distance h.
(c) The tearout area is
Atear  2  
Left pin
h L 
Right pin
h R 
Minimum value of h
RL
S s t
RR
S s t
d
h L  0.619  in
d
h R  0.804  in
h min  h R
h min  0.804  in
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-58-1
PROBLEM 4-58
_____
Statement:
Figure P4-20 shows a steel bar fastened to a rigid ground plane with two 0.25-in-dia hardened
steel dowel pins. For P = 2200 lb, find:
(a) The shear stress in each pin.
(b) The direct bearing stress in each pin and hole.
(c) The minimum value of dimension h to prevent tearout failure if the steel bar has a shear
strength of 32500 psi.
Given:
Pin diameter
Distance between pins
d  0.250  in
a  2.0 in
Thickness of bar
t  0.25 in
Solution:
1.
Applied load
Shear strength of bar
Distance from right pin to load
P  2200 lbf
S s  32.5 ksi
b  4.0 in
See Mathcad file P0458.
Draw a free-body diagram and find the shear forces (reactions) on each pin.
a
b
RL
h
RR
P
Write equations 3.3b for the bar and solve for the reactions.
 F:
RL 
2.
b
a
P
π d
2
4
RR  P  RL
RR  6600 lbf
2
A  0.0491 in
(a) Use equation 4.9 to determine the shear stress in each pin.
Left pin
Right pin
4.
RL  4400 lbf
RL  a  P b  0
Calculate the cross-section area of a pin.
A 
3.
 M:
RL  RR  P  0
τL 
τR 
RL
A
RR
A
τL  89.6 ksi
τR  134.5  ksi
(b) Calculate the bearing area from equation 4.10 and use it to determine the bearing stress in each pin.
Bearing area
Abear  d  t
2
Abear  0.0625 in
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
σL 
σR 
5.
(c) The tearout area is
Atear  2  
RL
Abear
RR
Abear
4-58-2
σL  70.4 ksi
σR  105.6  ksi
h  d

  t , where (h - d)/2 is the distance from the edge of the hole to
 2  
the outside of the bar. Substitute this area in equation 4.9 for the shear area and substitute the shear
strength for xy, solving then for the unknown distance h.
Left pin
h L 
Right pin
h R 
Minimum value of h
RL
S s t
RR
S s t
d
h L  0.792  in
d
h R  1.062  in
h min  h R
h min  1.062  in
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-59-1
PROBLEM 4-59
_____
Statement:
Figure P4-21 shows a rectangular section aluminum bar subjected to off-center forces P = 4000
N applied as shown.
(a) Solve for the maximum normal stress in the mid-region of the bar well away from the
eyes where the loads are applied.
(b) Plot the normal stress distribution across the cross section at this mid-region.
(c) Sketch a "reasonable" plot of the normal stress distribution across the cross section at
the ends, close to the applied loads.
Given:
Depth of bar
h  40 mm
Thickness of bar t  10 mm
Solution:
See Mathcad file P0459.
1.
Applied loads
Location of eye
P  4000 N
d  35 mm (from bottom edge)
Draw a free-body diagram of the bar, cut at any section along the length of the bar.
FACE OF CUT
SURFACE
P
d
P
M
0.5h
h
SECTION CENTROIDAL AXIS
Equilibrium requires that there be a force directed along the centroidal axis of the cross section that is equal
and opposite to the applied force and a bending moment to react the couple formed by the applied force and
the reaction force. Thus, since the reaction moment is clockwise,
M  ( d  0.5 h )  P
2.
Calculate the cross-section area, moment of inertia, and distance from the centroid to the outer surface.
A  h  t
I 
t h
A  400.0  mm
3
4
12
4
c  20.000 mm
(a) The normal stress on a section well away from the ends is a combination of uniform tension, as given by
equation 4.7, and bending, as given by equation 4.11a.
σ ( y )  
M y
I

P
A
This will be a maximum at y = c. σmax  σ ( c)
3.
2
I  5.333  10  mm
c  0.5 h
2.
M  60.000 N  m
σmax  32.5 MPa
(b) Plot the normal stress distribution across the cross section at the mid-region of the bar for
y  c c  1  mm  c
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-59-2
NORMAL STRESS ON SECTION
40
Stress, MPa
30
20
σ( y )
MPa
10
0
 10
 20
 20
 10
0
10
20
30
y
mm
Distance from neutral axis, mm
4.
(c) Sketch a "reasonable" plot of the normal stress distribution across the cross section at the ends, close to
the applied loads. Use the "force flow" analogy show in Figures 4-37 and 4-38 as a guide to the stress
distribution. Near the applied load the stress will be highly concentrated. As the distance from the point of
load application increases the stress will become more evenly distributed.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-60-1
PROBLEM 4-60
_____
Statement:
Figure P4-22 shows a bracket machined from 0.5-in-thick steel flat stock. It is rigidly attached to
a support and loaded with P = 5000 lb at point D. Find:
(a) The magnitude, location, and the plane orientation of the maximum normal stress at
section A-A.
(b) The magnitude, location, and the plane orientation of the maximum shear stress at
section A-A.
(c) The magnitude, location, and the plane orientation of the maximum normal stress at
section B-B.
(d) The magnitude, location, and the plane orientation of the maximum shear stress at
section B-B.
Given:
Distance from support to:
Section A-A
Point D d  8  in
Depth of section
h  3  in
Applied load
P  5000 lbf
Centroid of B-B b  18.5 in
a  10 in
Thickness of section
t  0.5 in
Assumptions: The bracket remains flat and does not buckle (out-of-plane) under the applied load.
Solution:
1.
See Mathcad file P0460.
Calculate the cross-section area and moment of inertia at sections A-A and B-B, which are the same.
2
A  h  t
2.
A  1.500  in
I 
t h
3
12
4
I  1.1250 in
For parts (a) and (b), draw a free-body diagram of the portion of the bracket that is to the right of section A-A.
V
a
A
M
y
h
A
d
x
D
P
3.
Use the equilibrium equations 3.3a to calculate the shear force and bending moment on section A-A.
 F:
V  P
4.
 M:
V  P  0
V  5000 lbf
P ( a  d )  M  0
M  P ( a  d )
M  10000  in lbf
(a) The maximum normal stress in the bracket at section A-A is determined using equation 4.11b. It is located a
the bottom of the section and is oriented in the positive x direction, i.e., it is tensile.
Distance from neutral axis to extreme fiber
c  0.5 h
c  1.500  in
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
σmax 
Maximum normal stress
5.
4-60-2
M c
σmax  13.33  ksi
I
(b) The maximum shear stress in the bracket at section A-A is either at the neutral axis (due to the transverse
shear, which is a maximum at the NA) or it is at the top or bottom of the section (due to the bending stress at
those points, which is numerically the same).
τmax 
At the neutral axis, using equation 4.14b
3 V

2 A
τmax  5.000  ksi
At the bottom edge the stress state is: σx  σmax, σy  0  ksi, τxy  0  ksi. Using equation 4.6a, the principal
stresses are
σ1 
σ2 
σx  σy
2
σx  σy
2
2
 σx  σy 
2
 
  τxy
 2 
σ1  13.333 ksi
2
 σx  σy 
2
 
  τxy
 2 
σ2  0.000  ksi
σ3  0  ksi
And, from equation 4.6b, the maximum shear stress is
τmax 
σ1  σ3
τmax  6.667  ksi
2
As seen from the Mohr's Circle in Figure 4-8, this stress is oriented 45 degrees from the positive x axis.
6.
For parts (c) and (d), draw a free-body diagram of the portion of the bracket that is below section B-B.
b
y
F
M
B
d
B
x
h
D
P
7.
Use the equilibrium equations 3.3a to calculate the normal force and bending moment on section B-B.
 F:
F  P  0
 M:
P ( b  d )  M  0
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
F  P
8.
4-60-3
F  5000 lbf
M  P ( b  d )
(c) The maximum normal stress in the bracket at section B-B is a combination of uniform tension and bending
and is determined by summing equations 4.7 and 4.11b. It is located at the left edge of the section and is
oriented in the positive y direction, i.e., it is tensile.
c  0.5 h
Distance from neutral axis to extreme fiber
Maximum normal stress
9.
M  52500  in lbf
σmax 
M c
I

F
A
c  1.500  in
σmax  73.33  ksi
(d) The maximum shear stress in the bracket at section B-B is at the left edge of the section (due to the
combined tensile and bending stresses). Since there is no transverse shear on this section, the shear stress
at the neutral axis is zero.
At the left edge the stress state is: σx  0  ksi, σy  σmax, τxy  0  ksi. Using equation 4.6a, the principal
stresses are
σ1 
σ2 
σx  σy
2
σx  σy
2
2
 σx  σy 
2
 
  τxy
2


σ1  73.333 ksi
2
 σx  σy 
2
 
  τxy
2


σ2  0.000  ksi
σ3  0  ksi
And, from equation 4.6b, the maximum shear stress is
τmax 
σ1  σ3
2
τmax  36.667 ksi
As seen from the Mohr's Circle in Figure 4-8, this stress is oriented 45 degrees from the positive x axis.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-61-1
PROBLEM 4-61
_____
Statement:
For the bracket of Problem 4-60, solve for the deflection of point C.
Given:
Distance from support to:
Point D
d  8  in
Depth of section h  3  in
Applied load
Point C
a  18.5 in
Thickness of section
t  0.5 in
P  5000 lbf
6
E  30 10  psi
Modulus of elasticity
Assumptions: 1. The bracket remains flat and does not buckle (out-of-plane) under the applied load.
2. The bracket can be modeled using its centroidal axis length dimensions.
Solution:
1.
See Mathcad file P0461.
Calculate the moment of inertia along the segment AC.
I 
2.
t h
3
4
I  1.1250 in
12
Draw idealized free-body diagrams of the portions of the bracket from the support to point C and from point
C to point D.
y
P
a
MA
MC
C
x
A
P
d
P
MC
C
D
P
3.
Calculate the magnitude of the moments on segment AC using equilibrium equation 3.3a.
MC  P ( a  d )
MC  52500  in lbf
MA  P d
MA  40000  in lbf
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4.
4-61-2
From inspection of the FBD, write the load function equation
q(x) = -MA<x - 0>-2 + P<x - 0>-1 - P<x - a>-1 - MC<x - a>-2
5.
Integrate this equation from - to x to obtain shear, V(x)
V(x) = -MA<x - 0>-1 + P<x - 0>0 - P<x - a>0 - MC<x - a>-1
6.
Integrate this equation from - to x to obtain moment, M(x)
M(x) = -MA<x - 0>0 + P<x - 0>1 - P<x - a>1 - MC<x - a>0
7.
Integrate the moment function, multiplying by 1/EI, to get the slope.
(x) = [-MA<x-0>1 + P<x - 0>2/2 - P<x - a>2/2 - MC<x-a>1 + C3]/EI
8.
Integrate again to get the deflection.
y(x) = [-MA<x-0>2/2 + P<x - 0>3/6 - P<x - a>3/6 - MC<x-a>2/2 + C3x +C4]/EI
9.
Evaluate C3 and C4. At x = 0,  = 0 and y = 0, therefore, C3 = 0 and C4 = 0.
10. Evaluate  and y at x = a using the equations in steps 7 and 8, respectively.
θC 
yC 
1
E I
  MA a 

P
2
a
2


 MA 2 P 3
a  a 
E I  2
6

1

θC  0.196  deg
yC  0.0465 in
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-62-1
PROBLEM 4-62
_____
Statement:
Figure P4-23 shows a 1-in-dia steel bar supported and subjected to the applied load P = 500 lb.
Solve for the deflection at the load and the slope at the roller support.
Given:
Diameter
Applied load
Solution:
See Mathcad file P0462.
1.
6
Modulus of elasticity
d  1.00 in
Dimensions:
P  500  lbf
a  20 in
E  30 10  psi
L  40 in
Draw a free-body diagram.
L
a
R2
M1
R1
2.
This is a statically indeterminate beam because there are three unknown reactions, R1, M1, and R2. To solve
for these unknowns, follow the method presented in Example 4-7. First, calculate the moment of inertia for
the round section.
I 
3.
P
π d
4
4
I  0.0491 in
64
From inspection of the FBD, write the load function equation
q(x) = -M1<x>-2 + R1<x>-1 - R2<x - a>-1 + P<x - L>-1
4.
Integrate this equation from - to x to obtain shear, V(x)
V(x) = -M1<x>-1 + R1<x>0 - R2<x - a>0 + P<x - L>0
5.
Integrate this equation from - to x to obtain moment, M(x)
M(x) = -M1<x>0 + R1<x>1 - R2<x - a>1 + P<x - L>1
6.
Integrate the moment function, multiplying by 1/EI, to get the slope.
(x) = [ -M1<x>1 + R1<x>2/2 - R2<x - a>2/2 + P<x - L>2/2 + C3]/EI
7.
Integrate again to get the deflection.
y(x) = [-M1<x>2/2 + R1<x>3/6 - R2<x - a>3/6 + P<x - L>3/6 + C3x + C4]/EI
8.
Evaluate R1, M1, R2, C3 and C4
At x = 0, y = 0 and  = 0, therefore, C3 = 0 and C4 = 0.
At x = a, y = 0
At x = L+, V = M = 0
Guess
M1  1000 in lbf
Given
y(a) = 0:
V(L) = 0:
R1  500  lbf

M1
2
2
a 
R1
6
3
R2  1000 lbf
3
 a = 0  lbf  in
R1  R2  P = 0  lbf
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
M1  R1 L  R2 ( L  a ) = 0  lbf  in
M(L) = 0:
 M1 
 
 R1   Find  M1 R1 R2
R 
 2
9.
4-62-2
M1  5000 in lbf
R1  750  lbf
R2  1250 lbf
Evaluate y at x = L to get the deflection at the load.
yL 
 M1 2 R1 3 R2
3
L 
L 
 ( L  a) 
E I  2
6
6

1

yL  1.584  in
10. Evaluate  at x = a to get the slope at the roller support.
θA 

E I 
1
  M1 a 
R1
2
a
2


θA  0.0340 rad
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-63-1
PROBLEM 4-63
_____
Statement:
Figure P4-24 shows a 1.25-in-dia solid steel shaft with several twisting couples applied in the
directions shown. For TA = 10000 lb-in, TB = 20000 lb-in, TC = 30000 lb-in, find:
(a) The magnitude and location of the maximum shear stress in the shaft.
(b) The corresponding principal stresses for the location determined in part (a).
(c) The magnitude and location of the maximum shear strain in the shaft.
Given:
Modulus of rigidity
Shaft diameter
d  1.25 in
Torque magnitudes:
TA  10 kip in TB  20 kip in
LAB  18 in
Segment lengths:
Solution:
1.
LBC  12 in
6
G  11.7 10  psi
TC  30 kip in
LCD  10 in
See Mathcad file P0463.
Looking at the shaft from the left end (A), TA and TC are clockwise (negative) and TB is counterclockwise
(positive). For the shaft to be in equilibrium, the applied torques must sum to zero. Write the equilibrium
equation and solve for the unknown reaction TD.
TA  TB  TC  TD  0
TD  TA  TB  TC
2.
3.
The net torque on each shaft segment is now
TAB  TA
TAB  10 kip in
TBC  TAB  TB
TBC  10 kip in
TCD  TBC  TC
TCD  20 kip in
Calculate the outside radius and the polar moment of inertia of the shaft.
r 
4.
d
r  0.625  in
2
J 
π d
4
4
J  0.240  in
32
(a) Since the shaft is uniform in cross-section, the maximum shear stress will occur in that segment that has
the largest absolute value of torque applied to it. In this case, that is segment CD. Use equation 4.23b to
calculate the maximum shear stress in segment CD.
τmax 
5.
TD  20 kip in
TCD  r
τmax  52.2 ksi
J
(b) Mohr's circle for pure shear is centered at 0,0 and has a radius equal to the shear stress on the stress
element. Thus, for this case, the two nonzero principal stresses are
σ1  τmax
σ1  52.2 ksi
σ3  τmax
σ3  52.2 ksi
The third principal stress is zero, σ2  0  ksi
6.
(c) The shear strain in any given segment is proportional to the shear stress so the maximum shear strain
will occur in segment CD, where the shear stress is a maximum. Hooke's law for shear is similar to that
given in equation 2.2.
γmax 
τmax
G
3
γmax  4.46  10
 rad
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-64-1
PROBLEM 4-64
_____
Statement:
If the shaft of Problem 4-63 were rigidly attached to fixed supports at each end (A and D) and
loaded only by the couples TB and TC, then find:
(a) The reactions TA and TD at each end of the shaft.
(b) The rotation of section B with respect to section C.
(c) The magnitude and location of the maximum shear strain.
Given:
Shaft diameter
Torque magnitudes:
d  1.25 in
TB  20 kip in
Modulus of rigidity
TC  30 kip in
Segment lengths:
LAB  18 in
LBC  12 in
Solution:
1.
LCD  10 in
See Mathcad file P0464.
Calculate the outside radius and the polar moment of inertia of the shaft.
r 
2.
6
G  11.7 10  psi
d
r  0.625  in
2
J 
π d
4
4
J  0.240  in
32
(a) Looking at the shaft from the left end (A), TC is clockwise (negative) and TB is counterclockwise
(positive). For the shaft to be in equilibrium, the applied torques must sum to zero. Since there are two
unknown reactions in the equilibrium equation, we cannot solve for them without another equation. An
equation that expresses the fact that the total rotational deflection from A to D is zero is called the
compatibility equation. Write the equilibrium and compatibility equations and solve for the unknown
reactions TA and TD.
TA  TB  TC  TD  0
θAB  θBC  θCD  0
TA  10 kip in
Guess
Given

TA  TB LBC
J G

TD LCD
J G
 0
TD  30 kip in
TA LAB
J G

TA  TB LBC
J G
 TA 
   Find  TA TD
 TD 
4.
J G

TA  TB  TC  TD = 0  kip in

3.
TA LAB

TD LCD
J G
= 0  rad
TA  3.50 kip in
clockwise
TD  13.50  kip in
counterclockwise
The net torque on each shaft segment is now
TAB  TA
TAB  3.5 kip in
TBC  TAB  TB
TBC  16.5 kip in
TCD  TBC  TC
TCD  13.5 kip in
(b) Use equation 4.24 to calculate the rotation of section B with respect to C.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
θBC  
5.
TBC LBC
J G
4-64-2
θBC  0.0706 rad
θBC  4.045  deg
(c) Since the shaft is uniform in cross-section, the maximum shear stress will occur in that segment that has the
largest absolute value of torque applied to it. In this case, that is segment BC. Use equation 4.23b to calculate
the maximum shear stress in segment BC.
τmax 
TBC  r
J
τmax  43.0 ksi
The shear strain in any given segment is proportional to the shear stress so the maximum shear strain will
occur in segment BC, where the shear stress is a maximum. Hooke's law for shear is similar to that given in
equation 2.2.
γmax 
τmax
G
3
γmax  3.68  10
 rad
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-65-1
PROBLEM 4-65
_____
Statement:
Figure P4-25 shows a pivot pin that is press-fit into part A and is slip fit in part B. If F = 100 lb
and l = 1.50 in, what pin diameter is needed to limit the maximum stress in the pin to 50 kpsi?
Given:
Applied force
Total length, l
F  100  lbf
l  1.50 in
Maximum stress σ  50 ksi
Beam length
L  0.5 l
Assumptions: 1. Since there is a slip fit between the pin and part B, part B offers no resistance to bending of
the pin and, since the pin is press-fit into part A, it can be modeled as a cantilever beam of
length l/2.
2. Part B distributes the concentrated force F so that, at the pin, it is uniformly distributed over
the exposed length of the pin.
Solution:
1.
See Mathcad file P0465.
Calculate the intensity of the uniformly distributed load acting over the length of the pin.
w 
2.
F
w  133.3 
L
lbf
in
A cantilever beam with uniform loading is shown in Figure B-1(b) in Appendix B. In this case, the dimension
a in the figure is zero. As shown in the figure, when a = 0, the maximum bending moment occurs at the
support and is
2
Mmax 
3.
w L
Mmax  37.50  lbf  in
2
The bending stress in a beam is given in equation 4.11c, which can be solved for the required section
modulus, Z.
Z 
Mmax
4
Z  7.500  10
σ
where, for a round cross-section
1
d min 
 32 Z 
 π 


z=
I
c
=
π d
3
 in
3
Solving for d,
32
3
d min  0.197  in
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
PROBLEM 4-66
4-66-1
_____
Statement:
Figure P4-25 shows a pivot pin that is press-fit into part A and is slip fit in part B. If F = 100 N
and l = 64 mm, what pin diameter is needed to limit the maximum stress in the pin to 250 MPa?
Given:
Applied force
Total length, l
F  100  N
l  64 mm
Maximum stress σ  250  MPa
Beam length
L  0.5 l
Assumptions: 1. Since there is a slip fit between the pin and part B, part B offers no resistance to bending of
the pin and, since the pin is press-fit into part A, it can be modeled as a cantilever beam of
length l/2.
2. Part B distributes the concentrated force F so that, at the pin, it is uniformly distributed over
the exposed length of the pin.
Solution:
1.
See Mathcad file P0466.
Calculate the intensity of the uniformly distributed load acting over the length of the pin.
w 
2.
F
w  3.125 
L
N
mm
A cantilever beam with uniform loading is shown in Figure B-1(b) in Appendix B. In this case, the dimension
a in the figure is zero. As shown in the figure, when a = 0, the maximum bending moment occurs at the
support and is
2
Mmax 
3.
w L
Mmax  1600.0 N  mm
2
The bending stress in a beam is given in equation 4.11c, which can be solved for the required section
modulus, Z.
Z 
Mmax
Z  6.400  mm
σ
where, for a round cross-section
z=
I
c
=
π d
3
3
Solving for d,
32
1
d min 
 32 Z 
 π 


3
d min  4.0 mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-67-1
PROBLEM 4-67
Statement:
Figure P4-25 shows a pivot pin that is press-fit into part A and is slip fit in part B. Determine
the l/d ratio that will make the pin equally strong in shear and bending if the shear strength is
equal to one-half the bending strength.
Assumptions: 1. Since there is a slip fit between the pin and part B, part B offers no resistance to bending of
the pin and, since the pin is press-fit into part A, it can be modeled as a cantilever beam of
length l/2.
2. Part B distributes the concentrated force F so that, at the pin, it is uniformly distributed
over the exposed length of the pin.
Solution:
1.
See Mathcad file P0467.
The intensity of the uniformly distributed load acting over the exposed length of the pin is
w 
2.
2 F
l
A cantilever beam with uniform loading is shown in Figure B-1(b) in Appendix B. In this case, the
dimension a in the figure is zero. As shown in the figure, when a = 0, the maximum bending moment for
a beam of length L occurs at the support and is
2
Mmax =
3.
=
2
M
Z
 F  l    32  = 8  F  l
 4   3
   π d  π d 3
=
Figure B-1(b) in Appendix B shows that the maximum shear occurs at the support and, for a = 0, is
Vmax = w L =
5.
 2 F    l  = F
 l  2

 
From equation 4.15c, the maximum shear stress due to the transverse loading is
τmax =
6.
1 2 F  l 
F l

  =
2 l 2
4
From equation 4.11c, the bending stress is
σmax =
4.
2
w L
4 V
4
4
16 F
 = F 
=
3 A
3
2
2
π d
3  π d
For equal shear and bending strength, let the shear stress equal one half the bending stress.
16 F
3  π d
Solving for l/d,
=
2
1 8 F  l

2
3
π d
l
d
=
4
3
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P0467.xmcd
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-69-1
PROBLEM 4-69
Statement:
Figure P4-26a shows a C-clamp with an elliptical body dimensioned as shown. The clamp has a
T-section with a uniform thickness of 3.2 mm at the throat as shown in Figure P4-26b. Find the
bending stress at the inner and outer fibers of the throat if the clamp force is 2.7 kN.
Given:
Clamping force F  2.7 kN
Distance from center of screw to throat
Section dimensions:
Solution:
1.
ri  63.5 mm
Flange b  28.4 mm
Determine the location of the CG of the T-section and the distance from the centerline of the screw to the
centroid of the section at the throat.
0.5 t ( b  t)  0.5 ( h  t)  ( h  t) t
yCG  9.58 mm
b t  ( h  t)  t
rc  ri  yCG
rc  73.08  mm
Using equation 4.12a and Figure 4-16, calculate the distance to the neutral axis, rn, and the distance from the
centroidal axis to the neutral axis, e.
Distance from the screw centerline to the outside fiber
Cross section area
Distance to neutral axis
A  b  t  ( h  t)  t
rn 
ri t
i
Distance from centroidal to neutral axis
A  182.4  mm
2
rn  71.86  mm
ro

t
dr  
dr
r
r

r  t
b
i
e  rc  rn
e  1.21 mm
M  rc F
5
M  1.97  10  N  mm
Calculate the distances from the neutral axis to the inner and outer fibers.
ci  rn  ri
5.
ro  95.30  mm
Take a section through the throat area and draw a FBD. There will be a vertical axial force through the
section CG (at a distance rc from the screw centerline) which will form a couple of magnitude rc x F. This
couple will be balanced by an internal moment of equal magnitude.
Internal moment
4.
ro  ri  h
A



r
3.
t  3.2 mm
See Figure P4-26 and Mathcad file P0469.
yCG 
2.
Web h  31.8 mm
ci  8.364  mm
co  ro  rn
co  23.436 mm
Using equations 4.12d and 4.12e, calculate the stresses at the inner and outer fibers of the throat section.
σi 
 ci  F

e A  ri  A
M
σo  

 co  F

e A  ro  A
M

σi  132.2  MPa
σo  204.3  MPa
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P0469.xmcd
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-70-1
PROBLEM 4-70
Statement:
A C-clamp as shown in Figure P4-26a has a rectangular cross section as in Figure P4-26c. Find
the bending stress at the inner and outer fibers of the throat if the clamping force is 1.6 kN.
Given:
Clamping force F  1.6 kN
Distance from center of screw to throat
Section dimensions:
Solution:
1.
Width b  6.2 mm
Depth h  31.8 mm
See Figure P4-26 and Mathcad file P0470.
Determine the distance from the centerline of the screw to the centroid of the section at the throat.
rc  ri 
2.
ri  63.5 mm
h
rc  79.40  mm
2
Using equation 4.12a and Figure 4-16, calculate the distance to the neutral axis, rn, and the distance from the
centroidal axis to the neutral axis, e.
Distance from the screw centerline to the outside fiber
Cross section area
Distance to neutral axis
A  b  h
rn 
A
ro
2
rn  78.33  mm
dr
i
e  rc  rn
e  1.07 mm
Take a section through the throat area and draw a FBD. There will be a vertical axial force through the
section CG (at a distance rc from the screw centerline) which will form a couple of magnitude rc x F. This
couple will be balanced by an internal moment of equal magnitude.
Internal moment
M  rc F
5
M  1.27  10  N  mm
Calculate the distances from the neutral axis to the inner and outer fibers.
ci  rn  ri
5.
b
r
Distance from centroidal to neutral axis
4.
ro  95.30  mm
A  197.160  mm



r
3.
ro  ri  h
ci  14.827 mm
co  ro  rn
co  16.973 mm
Using equations 4.12d and 4.12e, calculate the stresses at the inner and outer fibers of the throat section.
σi 
 ci  F

e A  ri  A
M
σo  

 co  F

e A  ro  A
M

σi  148.3  MPa
σo  98.8 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-71-1
PROBLEM 4-71
Statement:
Given:
A C-clamp as shown in Figure P4-26a has an elliptical cross section as in Figure P4-26d.
Dimensions of the major and minor axes of the ellipse are given. Determine the bending stress
at the inner and outer fibers of the throat if the clamping force is 1.6 kN.
Clamping force F  1.6 kN
Distance from center of screw to throat
ri  63.5 mm
Section dimensions:
Solution:
1.
Depth h  31.8 mm
See Figure P4-26 and Mathcad file P0471.
Determine the distance from the centerline of the screw to the centroid of the section at the throat.
rc  ri 
2.
Width b  9.6 mm
h
rc  79.40  mm
2
Using equation 4.12a and Figure 4-16, calculate the distance to the neutral axis, rn, and the distance from the
centroidal axis to the neutral axis, e.
ro  ri  h
Distance from the screw centerline to the outside fiber
Cross section area
Distance to neutral axis
b h
A  π 
2 2
A  239.766  mm
A
rn 
ro






r
2

 1   r  rc 
2 b
4
2 
h


dr
i
e  rc  rn
e  0.805  mm
M  rc F
5
M  1.27  10  N  mm
Calculate the distances from the neutral axis to the inner and outer fibers.
ci  rn  ri
5.
0.5
Take a section through the throat area and draw a FBD. There will be a vertical axial force through the
section CG (at a distance rc from the screw centerline) which will form a couple of magnitude rc x F. This
couple will be balanced by an internal moment of equal magnitude.
Internal moment
4.
2
rn  78.595 mm
r
Distance from centroidal to neutral axis
3.
ro  95.30  mm
ci  15.095 mm
co  ro  rn
co  16.705 mm
Using equations 4.12d and 4.12e, calculate the stresses at the inner and outer fibers of the throat section.
σi 
 ci  F

e A  ri  A
M
σo  

 co  F

e A  ro  A
M

σi  163.2  MPa
σo  108.7  MPa
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
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P0471.xmcd
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-72-1
PROBLEM 4-72
Statement:
Given:
A C-clamp as shown in Figure P4-26a has a trapezoidal cross section as in Figure P4-26e.
Determine the bending stress at the inner and outer fibers of the throat if the clamping force is
1.6 kN.
Clamping force F  1.6 kN
Distance from center of screw to throat
ri  63.5 mm
Section dimensions:
Solution:
1.
b o  3.2 mm
Determine the distance from the centerline of the screw to the centroid of the section at the throat.
h bi  2 bo

3 bi  bo
rc  76.75  mm
Using equation 4.12a and Figure 4-16, calculate the distance to the neutral axis, rn, and the distance from the
centroidal axis to the neutral axis, e.
Distance from the screw centerline to the outside fiber
Cross section area
Distance to neutral axis
A 
bi  bo
2
h
ro
bi 
bi  bo
h
2
rn  75.771 mm
  r  ri
dr
i
e  rc  rn
e  0.979  mm
Take a section through the throat area and draw a FBD. There will be a vertical axial force through the
section CG (at a distance rc from the screw centerline) which will form a couple of magnitude rc x F. This
couple will be balanced by an internal moment of equal magnitude.
Internal moment
M  rc F
5
M  1.228  10  N  mm
Calculate the distances from the neutral axis to the inner and outer fibers.
ci  rn  ri
5.
ro  95.30  mm
A  203.520  mm
r
Distance from centroidal to neutral axis
4.
ro  ri  h
A
rn 





r
3.
Depth h  31.8 mm
See Figure P4-26 and Mathcad file P0472.
rc  ri 
2.
Width b i  9.6 mm
ci  12.271 mm
co  ro  rn
co  19.529 mm
Using equations 4.12d and 4.12e, calculate the stresses at the inner and outer fibers of the throat section.
σi 
 ci  F

e A  ri  A
M
σo  

 co  F

e A  ro  A
M

σi  126.9  MPa
σo  118.4  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,
P0472.xmcd
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-73-1
PROBLEM 4-73
Statement:
We want to design a C-clamp with a T-section similar to the one shown in Figure P4-26. The
depth of the section will be 31.8 mm as shown but the width of the flange (shown as 28.4 mm)
is to be determined. Assuming a uniform thickness of 3.2 mm and a factor of safety against
static yielding of 2, determine a suitable value for the width of the flange if the C-clamp is to be
made from 60-40-18 ductile iron and the maximum design load is 1.6 kN.
Given:
Maximum clamping force F  1.6 kN
Distance from center of screw to throat
ri  63.5 mm
Section dimensions:
Web h  31.8 mm
Factor of safety N  2 Yield strength
S y  324  MPa
Solution:
1.
t  3.2 mm
See Figure P4-26 and Mathcad file P0473.
Determine the location of the CG of the T-section and the distance from the centerline of the screw to the
centroid of the section at the throat as functions of the unknown flange width, b.
yCG ( b ) 
0.5 t ( b  t)  0.5 ( h  t)  ( h  t) t
b  t  ( h  t)  t
rc( b )  ri  yCG ( b )
2.
Using equation 4.12a and Figure 4-16, calculate the distance to the neutral axis, rn, and the distance from the
centroidal axis to the neutral axis, e,as functions of b.
ro  ri  h
Distance from the screw centerline to the outside fiber
A ( b )  b  t  ( h  t)  t
Cross section area
Distance to neutral axis
A (b)
rn( b ) 
r t
 i


r
i
Distance from centroidal to neutral axis
3.
e( b )  rc( b )  rn( b )
Calculate the distances from the neutral axis to the inner and outer fibers.
co( b )  ro  rn( b )
Using equations 4.12d and 4.12e, calculate the stresses at the inner and outer fibers of the throat section.
σi( b ) 
6.
i
M ( b )  rc( b )  F
ci( b )  rn( b )  ri
5.
r
 o t
dr  
dr
r
r

r t
b
Take a section through the throat area and draw a FBD. There will be a vertical axial force through the
section CG (at a distance rc from the screw centerline) which will form a couple of magnitude rc x F. This
couple will be balanced by an internal moment of equal magnitude.
Internal moment
4.
ro  95.3 mm
 ci( b ) 
F


e( b )  A ( b )  ri  A ( b )
M (b)

Set the tensile stress on the inner fiber equal to the yield strength divided by the factor of safety and solve
for the flange width, b.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
7.
Guess
b  12 mm
Given
σi( b ) =
Sy
N
b  Find ( b )
4-73-2
b  10.13  mm
Using the calculated value of b, check the stresses at the inner and outer fibers..
σi( b ) 
 ci( b ) 
F


e( b )  A ( b )  ri  A ( b )
σo( b )  
M (b)

 c o( b ) 
F

e( b )  A ( b )  ro  A ( b )
M ( b)

σi( b )  162  MPa
σo( b )  149.4  MPa
A suitable minimum value for the flange width is
b  10.1 mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-74-1
PROBLEM 4-74
Statement:
A round steel bar is 10 in long and has a diameter of 1 in.
(a) Calculate the stress in the bar when it is subjected to a 1000-lb force in tension.
(b) Calculate the bending stress in the bar if it is fixed at one end (as a cantilever beam) and
has a 1000-lb transverse load at the other end.
(c) Calculate the transverse shear stress in the bar of part (b).
(d) Calculate the torsional shear stress when the 1000-lb force is displaced 10 inches radially
from the centerline (axis) of the cantilever beam.
(e) Calculate the maximum bending stress in the bar if it is formed into a semicircle with a
centroidal radius of 10/ in and 1000-lb opposing forces are applied at the ends in the plane of
the of the ends. Assume that there is no distortion of the cross section during bending.
(f) Calculate the direct bearing stress that would result on the bar of (a) if it were the pin in a
pin-and-clevis connection that is subjected to a 1000-lb pull if the center part (the eye or
tongue) is 1-in wide.
(g) Determine how short the bar must be when loaded as a cantilever beam for its maximum
flexural bending stress and its maximum transverse shear stress to provide equal tendency to
failure. Find the length as a fraction of the diameter if the failure stress in shear is half the
failure stress in bending.
(h) If the force on the cantilever beam in (f) is eccentric, inducing torsional as well as bending
stress, what fraction of the diameter would the eccentricity need to be in order to give a
torsional stress equal to the transverse shear stress?
Given:
Length of bar
L  10 in
Force
F  1000 lbf
Solution:
See Mathcad file P0474.
Diameter
Load Radius
d  1.00 in
R  10 in
(a) Use equation 4.7 to calculate the axial stress.
Cross sectional area
Axial stress
A  π
σ 
d
2
2
A  0.785  in
4
F
σ  1.27 ksi
A
(b) The beam loading diagram is shown in Appendix Figure B-1a with the concentrated load at a = L. The
maximum bending stress occurs at x = 0 and is given by Equation 4.11b.
Bending moment
M  L F
Radius of bar
c 
Moment of inertia
Maximum bending stress
M  10000  in lbf
d
c  0.5 in
2
I  π
d
4
4
I  0.049  in
64
σ 
M c
I
σ  101.9  ksi
(c) The maximum transverse shear stress occurs at y = 0 and is given by Equation 4.15c and in Figure 4-20b.
Maximum transverse shear stress in a solid, round bar
τmax 
4 F

3 A
τmax  1.70 ksi
(d) The maximum torsional shear shear stress occurs at y = d/2 and is given by Equation 4.23b.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-74-2
Twisting torque
T  F  R
Polar moment of inertia
J  π
τ 
Max torsional shear stress
d
T  10000  in lbf
4
4
J  0.098  in
32
Td
τ  50.93  ksi
2 J
(e) The maximum bending stress for a curved beam occurs at r = ri and is given by Equation 4.12d.
10
Centroidal radius
rc 
Inside radius
ri  rc  0.5 d
ri  2.683  in
Outside radius
ro  rc  0.5 d
ro  3.683  in
Cross section area
A  π
rn 
Distance to neutral axis
 in
π
d
rc  3.183  in
2
2
A  0.785  in
4
A
rn  3.163  in
r
 o

 2


r
2
 d   r  r 2
 
c
2
dr
r
i
Distance from centroid to neutral axis
e  rc  rn
M  rc F
Internal moment
e  0.020  in
M  3183 in lbf
Distances from the neutral axis to the inner and outer fibers
ci  rn  ri
ci  0.480  in
co  ro  rn
co  0.520  in
Stress at the inner fibers of the throat section
σi 
 ci  F

e A  ri  A
M

σi  37.9 ksi
(f) The direct bearing stress is given in Equations 4.7 and 4.10.
Given length of bearing contact
l  1  in
Projected area of contact
Abearing  l d
Bearing stress
σbearing 
2
Abearing  1  in
F
Abearing
σbearing  1.0 ksi
(g) Determine how short the bar must be when loaded as a cantilever beam for its maximum flexural bending
stress and its maximum transverse shear stress to provide equal tendency to failure. Find the length as a
fraction of the diameter if the failure stress in shear is half the failure stress in bending.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
Bending stress
M c
σ=
I
=
4-74-3
32 F  L
π d
3
4 F
16 F
 =
3 A
2
3  π d
Transverse shear
τ=
Equating
σ = 2 τ
32 F  L
Solving for L
L=
π d
=
3
32
3

F
π d
2
d
3
(h) If the force on the cantilever beam in (f) is eccentric, inducing torsional as well as bending stress, what
fraction of the diameter would the eccentricity need to be in order to give a torsional stress equal to the
transverse shear stress?
Torsional shear stress
Transverse shear
Equating
τtor =
Tc
τtrans =
J
=
16 F  e
π d
3
4 F
16 F
 =
3 A
2
3  π d
16 F  e
τtor = τtrans
π d
Solving for the eccentricity, e
e=
3
=
16 F
3  π d
2
d
3
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-75a-1
PROBLEM 4-75a
Statement:
For a filleted flat bar in tension similar to that shown in Appendix Figure C-9 and the data from
row a from Table P4-4, determine the nominal stress, the geometric stress concentration factor,
and the maximum axial stress in the bar.
Given:
Widths
Thickness
Force
Solution:
See Appendix Figure C-9 and Mathcad file P0475a.
1.
P
σnom  40.0 MPa
h d
Determine the geometric stress concentration factor using Appendix Figure C-9.
Width ratio
D
d
3.
d  20 mm
Radius r  4  mm
Determine the nominal stress in the bar using equation 4.7.
σnom 
2.
D  40 mm
h  10 mm
P  8000 N
 2.00
From Figure E-9
A  1.0966
SCF
Kt  A  
r
b  0.32077
b

d
Kt  1.838
Determine the maximum stress in the bar using equation 4.31.
σmax  Kt σnom
σmax  73.5 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-76a-1
PROBLEM 4-76a
Statement:
For a filleted flat bar in bending similar to that shown in Appendix Figure C-10 and the data from
row a from Table P4-4, determine the nominal stress, the geometric stress concentration factor,
and the maximum bending stress in the bar.
Given:
Widths
Thickness
Moment
Solution:
See Appendix Figure C-10 and Mathcad file P0476a.
1.
d
c  10 mm
2
σnom 
M c
I 
h d
3
3
I  6.667  10  mm
12
4
σnom  120.0  MPa
I
Determine the geometric stress concentration factor using Appendix Figure C-10.
Width ratio
D
d
3.
d  20 mm
Radius r  4  mm
Determine the nominal stress in the bar using equation 4.11b.
c 
2.
D  40 mm
h  10 mm
M  80 N  m
 2.00
From Figure E-9
A  0.93232
SCF
Kt  A  
r
b  0.30304
b

d
Kt  1.518
Determine the maximum stress in the bar using equation 4.31.
σmax  Kt σnom
σmax  182.2  MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-77a-1
PROBLEM 4-77a
Statement:
For a shaft, with a shoulder fillet, in tension similar to that shown in Appendix Figure C-1 and
the data from row a from Table P4-4, determine the nominal stress, the geometric stress
concentration factor, and the maximum axial stress in the shaft.
Given:
Widths
Radius
Force
Solution:
See Appendix Figure C-1 and Mathcad file P0477a.
1.
D  40 mm
r  4  mm
P  8000 N
Determine the nominal stress in the bar using equation 4.7.
σnom 
4 P
π d
2.
σnom  25.5 MPa
2
Determine the geometric stress concentration factor using Appendix Figure C-1.
Width ratio
D
d
3.
d  20 mm
 2.00
From Figure E-1
A  1.01470
SCF
Kt  A  
r
b  0.30035
b

d
Kt  1.645
Determine the maximum stress in the bar using equation 4.31.
σmax  Kt σnom
σmax  41.9 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-78a-1
PROBLEM 4-78a
Statement:
For a shaft, with a shoulder fillet, in bending similar to that shown in Appendix Figure C-2 and
the data from row a from Table P4-4, determine the nominal stress, the geometric stress
concentration factor, and the maximum bending stress in the shaft.
Given:
Widths
Radius
Moment
Solution:
See Appendix Figure C-2 and Mathcad file P0478a.
1.
d
c  10 mm
2
σnom 
M c
I 
π d
4
3
I  7.854  10  mm
64
4
σnom  101.9  MPa
I
Determine the geometric stress concentration factor using Appendix Figure C-2.
Width ratio
D
d
3.
d  20 mm
Determine the nominal stress in the bar using equation 4.11b.
c 
2.
D  40 mm
r  4  mm
M  80 N  m
 2.00
From Figure E-2
A  0.90879
SCF
Kt  A  
r
b  0.28598
b

d
Kt  1.44
Determine the maximum stress in the bar using equation 4.31.
σmax  Kt σnom
σmax  146.7  MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-79-1
PROBLEM 4-79
Statement:
A differential stress element has a set of applied stresses on it as shown in Figure 4-1. For σx =
850, σy = -200, σz = 300, τxy = 450, τyz = -300, and τzx = 0; find the principal stresses and maximum
shear stress and draw the Mohr's circle diagram for this three-dimensional stress state.
Given:
σx  850
σy  200
σz  300
τxy  450
τyz  300
τzx  0
See Figure 4-1 and Mathcad file P04079.
Solution:
1. Calculate the coefficients (stress invariants) of equation (4.4c).
C2  σx  σy  σz
C1 
C2  950.000
 σx τxy 
 σx τzx 
 σy τyz 

  
  

 τxy σy 
 τzx σz 
 τyz σz 
 σx τxy τzx 


C0   τxy σy τyz 
τ τ σ 
 zx yz z 
3
σ2  r
σ2  388
σ3  r
σ3  470
3
2
1
 CW
 1-3
500
 1-2
 2-3
-500
4. Using equations (4.5), evaluate
the principal shear stresses.
τ13 
τ12 
τ23 
σ1  σ3
2
σ1  σ2
2
σ2  σ3
2
8
 470 
r   388 


 1032 
3. Extract the principal stresses from
the vector r by inspection.
σ1  1032
C0  1.882  10
2
r  polyroots ( v)
σ1  r
5
σ  C2 σ  C1 σ  C0 = 0
2. Find the roots of the triaxial stress equation:
 C0 


C1 

v 
 C2 


 1 
C1  2.675  10
3
500
1000
2
0
1
1500

τ13  751
500
τ12  322
τ23  429
5. Draw the three-circle Mohr diagram.
 CCW
FIGURE 4-79
The Three Mohr's Circles for Problem 4-79
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-80-1
PROBLEM 4 - 80
Statement:
Write expressions for the normalized (stress/pressure) tangential stress as a function of the
normalized wall thickness (wall thickness/outside radius) at the inside wall of a thick-wall
cylinder and for a thin-wall cylinder, both with internal pressure only. Plot the ratio of these two
expressions and determine the range of the wall thickness to outside radius-ratio for which the
stress predicted by the thin-wall expression is at least 5% greater than that predicted by the
thick-wall expression.
Solution:
See Mathcad file P0480.
1.
Let the σt/p ratio be S' and the t/ro ratio be t', then
For the thick-wall cylinder at the inside wall, using equation 4.48a
2
S'thick ( t') 
2  2  t'  t'
2
2  t'  t'
and, for the thin-wall cylinder, using equation 4.49a
S'thin ( t') 
1
t'
2.
Choose a range for the normalized thickness ratio, t'  0.01 0.02  0.99
3.
Plot the difference between the two functions. Δ ( t') 
S'thin( t')  S'thick( t') 
S'thick ( t')
25
20
15
Δ ( t')
%
10
5
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
t'
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P0480.xmcd
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4.
Determine the values of t' for which the difference is 5%.
Δ ( 0.10)  5.0 %
5.
4-80-2
Δ ( 0.946 )  5.1 %
The range of the normalized thickness for which the thin-wall stress is at least 5% greater than the thick-wall
stress is from 0.10 to 0.946.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-81-1
PROBLEM 4 - 81
Statement:
A hollow square torsion bar such as that shown in Table 4-3 has dimensions a = 25 mm, t = 3
mm, and l = 300 mm. If it is made of steel with a modulus of rigidity of G = 80.8 GPa, determine
the maximum shear stress in the bar and the angular deflection under a torsional load of 500
N-m.
Given:
Dimensions
Modulus
Solution:
See Table 4-3 Mathcad file P0481.
1.
a  25 mm
G  80.8 GPa
2
2  t  ( a  t)
2 a t  2 t
Q  2  t ( a  t)
4
2
2
K  31944  mm
Q  2904 mm
4
3
Using equation 4.26a, calculate the maximum shear stress.
τmax 
3.
l  300  mm
T  500  N  m
Calculate the factors K and Q for a hollow square from Table 4-3.
K 
2.
t  3  mm
Load
T
Q
τmax  172.2  MPa
Using equation 4.26b, calculate the angular deflection.
θ 
Tl
K G
θ  0.058 radians
θ  3.33 deg
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-82-1
PROBLEM 4 - 82
Statement:
Design a hollow rectangular torsion bar such as that shown in Table 4-3 that has dimensions a
= 45 mm, b = 20 mm, and l = 500 mm. It is made of steel with a shear yield strength of 90 MPa
and has an applied torsional load of 135 N-m. Use a factor of safety against yielding of 2.
Given:
Dimensions
a  45 mm
b  20 mm
l  500  mm
Modulus
G  80.8 GPa Load
T  135  N  m
Shear yield strength
S sy  90 MPa Factor of safety N  2
Solution:
See Table 4-3 Mathcad file P0482.
1.
Calculate the Q-factor for a hollow rectangle from Table 4-3.
Q( t)  2  t ( a  t)  ( b  t)
2.
Calculate the maximum shear stress as a function of thickness, t, using equation 4.26a.
τ( t) 
3.
T
Q( t)
Define a function that relates the maximum shear stress to the shear strength divided by the factor of safety
and solve for the thickness, t.
Guess a value of t
t  3  mm
Define the design function
f ( t)  τ( t) 
t  root( f ( t) t)
S sy
N
t  1.927  mm
Let t = 2 mm (note that this solution does not check for buckling under the applied load)
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-83-1
PROBLEM 4 - 83
Statement:
A pressure vessel with closed ends has the following dimensions: outside diameter, OD = 450
mm, and wall thickness, t = 6 mm. If the internal pressure is 690 kPa, find the principal stresses
on the inside surface away from the ends. What is the maximum shear stress at the point
analyzed?
Given:
Dimensions
Pressure
Solution:
See Mathcad file P0483.
1.
2.
OD  450  mm
p  690  kPa
t  6  mm
Convert the given dimensions to inside and outside radii.
ro  0.5 OD
ro  225  mm
ri  ro  t
ri  219  mm
Determine whether to use thick-wall or thin-wall theory.
ro
10
 22.5 mm
Since the wall thickness, t  6  mm, is much less than one tenth the outside radius, use thin wall theory.
3.
Calculate the principal stresses using equations 4.49.
Tangential (y-direction)
σt 
p  ro
t
Radial (x-direction)
σr  0  MPa
Axial (z-direction)
σa 
p  ro
2 t
σt  25.9 MPa
σr  0.0 MPa
σa  12.9 MPa
The principal stresses are:
4.
σ1  σt
σ1  25.9 MPa
σ2  σa
σ2  12.9 MPa
σ3  σr
σ3  0.0 MPa
Using equation 4.6b, calculate the maximum shear stress.
τmax 
σ1  σ3
2
τmax  12.9 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-84-1
PROBLEM 4-84
Statement:
A simply supported steel beam of length, l, with a concentrated load, F, acting at midspan has a
rectangular cross-section with width, b, and depth, h. If the strain energy due to transverse
shear loading is Us and that due to bending loading is Ub, derive an expression for the ratio
Us/Ub and plot it as a function of h/l over the range 0 to 0.10.
Solution:
See Mathcad file P0484.
l
1.
From equation 4.22e, the strain energy in transverse loading is:

2
3  V
Us =  
dx
5  G A

0
l
2.
From equation 4.22d, the strain energy in bending loading is:

2
1  M
Ub =  
dx
2  E I

0
l
3.
Let
U' =
Us
Ub
 2
 V dx

U' =
, then:
0
6 E I


5 G A l

2
 M dx

0
4.
For a rectangular cross-section:
A = b h
I=
and
b h
3
12
l
5.
E
And, for steel:
G
=
5
2
 2
 V dx
2 
0
h
U' =

l
4

2
 M dx

therefore
0
6.
7.
For the given loading:
F
For x between 0 and l/2,
V=
For x between l/2 and l,
V=
and
2
F
2
and
M=
M=
F x
2

F x
2
F l
2
Substituting these expressions into the equation for U' and integrating gives:
l

  0.5 l


2
2


F
F
  dx
   dx


12.0 

2 12.0
2
 2
h 








2  
2
2 
2


h  0
0.5 l
l
l 
6.0 h






4
4  0.5 l
l
2



l
2
2 


F

x
F

x
F

l
  dx 



 2 
 2  2  dx

 

 



0.5 l
 0

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MACHINE DESIGN - An Integrated Approach, 4th Ed.
h' =
h
4-84-2
2
U'( h')  6  h'
then
8.
Let
9.
Plotting the strain energy ratio over the range:
l
h'  0 0.001  0.10
STRAIN ENERGY RATIO vs DEPTH TO LENGTH RATIO
6
Strain Energy Ratio - Percent
5
4
U'( h' )
%
3
2
1
0
0
0.02
0.04
0.06
0.08
0.1
h'
Depth to Length Ratio
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-85a-1
PROBLEM 4-85a
Statement:
A beam is supported and loaded as shown in Figure P4-27(a). Find the reactions for the data
given in row a from Table P4-2.
Given:
Beam length
Distance to R2
L  1  m
a  0.4 m
Distributed load magnitude
w  200  N  m
1
L
a
w
R1
R2
R3
FIGURE 4-85A
Free Body Diagram for Problem 4-85
Solution:
See Figure P4-27(a) and Mathcad file P0485a.
1.
Consider the reaction force R1 to be redundant and remove it temporarily. The beam will then be statically
determinant and will deflect at x = 0. Now consider the reaction force R1 to be an unkown applied load that
will force the deflection to be zero. Write an equation for the deflection at x = 0 in terms of the force R1 with
the deflection set to zero.
2.
Write equation 4.21 for the deflection y1 at the unknown applied load R1 in terms of the strain energy in
y1 =
the beam at that point:
3.
Substitute equation 4.22d and differentiate:


y1 = 



U
R1
L
M
E I


 R1

M  dx
(a)

0
4.
5.
Write an expression for the bending moment and its partial derivative with respect toR1 as a function of x.
For x between 0 and a,
M = R 1 x 
For x between a and l,
M = R 1 x 
w x
2
2
w x
2
2
 R 2 ( x  a )

M =x
R1
(b)

M =x
R1
(c)
Substitute equations (b) and (c) into (a), set equal to zero and integrate.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.





a
2

 R1 x  w x   x dx 
2 

0
6.





L
4-85a-2
2


R1 x  w x  R2 ( x  a)  x dx = 0
2


a
Solving for R1 and R2 and summing forces and moments about x = 0:
From strain energy
R1
3
3
m 
4
 L3 a  L2 a 3 a3 
 
  R2  w  L = 0


2
8
3
2 
3
Summing forces
R1  R2  R3  w  L = 0
Summing moments
R 2 a  R 3 L 
2
7.
w L
=0
2
R1  65 N
Use these three equations to solve for R1, R2, and R3. Guess
Given
R1
3
3
m 
R2  70 N
R3  65 N
4
 L3 a  L2 a 3 a3 
 
  R2  w  L = 0


2
8
3
2 
3
R1  R2  R3  w  L = 0
2
R 2 a  R 3 L 
w L
=0
2
 10.714 
R   148.81   N


 40.476 
R  Find  R1 R2 R3
R  10.7 N
1
R  148.8  N
2
R  40.5 N
3
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-86a-1
PROBLEM 4-86a
Statement:
A beam is supported and loaded as shown in Figure P4-27(b). Find the reactions for the data
given in row a from Table P4-2.
Given:
Beam length
Distance to R2
L  1.0 m
a  0.4 m
Distributed load magnitude
Distance to concentrated load
Concentrated load
w  200  N  m
b  0.6 m
F  500  N
1
L
b
a
F
w
R1
R2
R3
FIGURE 4-86A
Free Body Diagram for Problem 4-86
Solution:
See Figure P4-27(b) and Mathcad file P0485a.
1.
Consider the reaction force R1 to be redundant and remove it temporarily. The beam will then be statically
determinant and will deflect at x = 0. Now consider the reaction force R1 to be an unkown applied load that
will force the deflection to be zero. Write an equation for the deflection at x = 0 in terms of the force R1 with
the deflection set to zero.
2.
Write equation 4.21 for the deflection y1 at the unknown applied load R1 in terms of the strain energy in
y1 =
the beam at that point:
3.
Substitute equation 4.22d and differentiate:


y1 = 



U
R1
L
M
E I



M  dx
(a)
 R1 
0
4.
Write an expression for the bending moment and its partial derivative with respect toR1 as a function of x.
w x
2
For x between 0 and a,
M = R 1 x 
For x between a and b,
M = R1 x  w a   x 
2

a
  R 2 ( x  a )
2

M =x
R1
(b)

M =x
R1
(c)
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4-86a-2
M = R1 x  w a   x 
For x between b and L,

a
  R 2 ( x  a )  F  ( x  b )
2
(d)

M =x
R1
5.
Substitute equations (b), (c) and (d) into (a), set equal to zero and integrate.





a
2

 R1 x  w x   x dx 
2 

0




b
R  x  w a   x 
 1



a

  R2 ( x  a )  x dx  = 0

2
a




L
R  x  w a   x 
 1



a

  R2 ( x  a )  F  ( x  b )  x dx

2
b
6.
Solving for R1 and R2 and summing forces and moments about x = 0:
From strain energy
7.
2
3
 L3 
 3
   R1   L  a  L  a   R2 
=0
2
6 
3
3
2
3
 a2 L2 a L3 a4 
 3
  w   L  b L  b   F



3
2
24 
6 
 6
3
Summing forces
R1  R2  R3  w  a  F = 0
Summing moments
R 2 a  R 3 L 
w a
2
 F b = 0
2
Use these three equations to solve for R1, R2, and R3. Guess R1  100  N
Given
R2  400  N
R3  200  N
2
3
 L3 
 3
   R1   L  a  L  a   R2 
=0
2
6 
3
3
 a2 L2 a L3 a4 
 L3 b  L2 b 3 


F



w  


3
2
24 
6 
 6
3
R1  R2  R3  w  a  F = 0
R 2 a  R 3 L 
w a
2
 F b = 0
2
 81.143 
R   575.238   N


 85.905 
R  Find  R1 R2 R3
R  81.1 N
1
R  575.2  N
2
R  85.9 N
3
© 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,
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Pearson Education, Inc., Upper Saddle River, NJ 07458.
MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-1a-1
PROBLEM 5-1a
Statement:
A differential stress element has a set of applied stresses on it as indicated in row a of Table P5-1.
For row a, draw the stress element showing the applied stresses. Find the principal stresses and
the von Mises stress.
Given:
σx  1000
σy  0
σz  0
τxy  500
τyz  0
τzx  0
Solution:
See Figure 5-1a and Mathcad file P0501a.
1. Draw the stress element, indicating the x and y axes.
500
2. From Problem 4-1a, the principal stresses are
σ1  1207
σ2  0
y
σ3  207
x
1000
3. Using equatoion 5.7c, the von Mises stress is
σ' 
2
σ1  σ1 σ3  σ3
2
σ'  1323
FIGURE 5-1aA
Stress Element for Problem 5-1a
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-1h-1
PROBLEM 5-1h
Statement:
A differential stress element has a set of applied stresses on it as indicated in each row of Table
P5-1. For row h, draw the stress element showing the applied stresses, find the principal stresses
and the von Mises stress.
Given:
σx  750
σy  500
σz  250
τxy  500
τyz  0
τzx  0
Solution:
See Figure 5-1h and Mathcad file P0501h.
z
1. Draw the stress element (see Figure 5-1h).
250
2. From Problem 4-1h, the principal stresses are
σ1  1140
σ2  250
σ3  110
3. Using equation 5.7, the von Mises stress is
σ' 
1
2
σ'  968
  σ1  σ2   σ2  σ3   σ1  σ3
2
2
750
2

500
500
x
500
y
FIGURE 5-1h
Stress Element for Problem 5-1h
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-2-1
PROBLEM 5-2
Statement:
A 400-lb chandelier is to be hung from two 10-ft-long solid, low- carbon steel cables in tension.
Size the cables for a safety factor of 4. State all assumptions.
Given:
Weight of chandelier
Length of cable
Design Safety factor
W  400  lbf
L  10 ft
Nd  4
Number of cables
N  2
Young's modulus
E  30 10  psi
L  120 in
6
Assumptions: The material is AISI 1010 hot-rolled steel with S y  26 ksi
Solution:
See Mathcad file P0502.
P 
W
P  200 lbf
1.
Determine the load on each cable
2.
Using the distortion-energy failure theory,
3.
In this case, the only stress in the axial direction is the tensile stress. Therefore, this is the principal stress and
also the von Mises stress.
4 P
σ' = σ1 = σ =
2
π d
4.
Substitute the equation in step 3 into the design equation in step 2 and solve for the minimum diameter, d.
N
Nd =
Sy
σ'
1
 4  P N d 
d  

 π S y 
5.
2
d  0.198 in
Round up to an available size (see Table 13-2) and check the actual factor of safety against static failure.
2
d  0.207  in
Ns 
π d  S y
4 P
Ns  4.4
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-3-1
PROBLEM 5-3
Statement:
For the bicycle pedal-arm assembly in Figure P5-1 with rider-applied force of 1500 N at the pedal,
determine the von Mises stress in the 15-mm-dia pedal arm. The pedal attaches to the arm with a
12-mm thread. Find the von Mises stress in the screw. Find the safety factor against static failure
if the material has S y = 350 MPa.
Given:
Distances (see figure)
Rider-applied force
a  170  mm
Frider  1.5 kN
b  60 mm
Screw thread diameter
d sc  12 mm
Pedal arm diameter
d pa  15 mm
Material yield strength
S y  350  MPa
See Figures 5-3 and Mathcad file P0503.
z
Solution:
1. From problem 4-3, the maximum principal
stresses in the pedal arm are at point A and
are
σ1  793  MPa
a
σ2  0  MPa
σ' 
2
Mc
b
σ3  23 MPa
2. Using equation 5.7c, the von Mises
stress is
σ1  σ1 σ3  σ3
Tc
C
Frider
Arm
y
Fc
Pedal
2
x
FIGURE 5-3A
σ'  805 MPa
Free Body Diagram for Problem 5-3
3. The factor of safety for the pedal arm is
N 
Sy
σ'
z
N  0.43
4. From Problem 4-3 solution, the stresses
at the top of the screw where it joins the
pedal arm are
σx  530.5  MPa
Section C
A
σz  0  MPa
B
Arm
τzx  0  MPa
x
5. From this, we see that the principal
stresses are
σ1  σx
y
σ2  0  MPa
FIGURE 5-3B
σ3  0  MPa
Points A and B at Section C
6. The von Mises stress is
σ'  σ1
7. The factor of safety for the screw is
N 
Sy
σ'
σ'  530.5 MPa
N  0.66
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-4-1
PROBLEM 5-4
Statement:
The trailer hitch shown in Figure P4-2 and Figure 1-1 (p. 12) has loads applied as defined in
Problem 3-4. The tongue weight of 100 kg acts downward and the pull force of 4905 N acts
horizontally. Using the dimensions of the ball bracket shown in Figure 1-5 (p. 15) and S y = 300
MPa ductile steel, determine static safety factors for:
(a) The shank of the ball where it joins the ball bracket.
(b) Bearing failure in the ball bracket hole.
(c) Tearout failure in the ball bracket.
(d) Tensile failure in the 19-mm diameter attachment holes.
(e) Bending failure in the ball bracket as a cantilever.
Given:
a  40 mm
b  31 mm
c  70 mm
d  20 mm
Mtongue  100  kg Fpull  4.905  kN
d sh  26 mm
t  19 mm
S y  300  MPa
Assumptions: 1. The nuts are just snug-tight (no pre-load), which is the worst case.
2. All reactions will be concentrated loads rather than distributed loads or pressures.
Solution:
See Figure 5-4 and Mathcad file P0504.
W tongue
70 = c
1
F pull
1
40 = a
2
A
B
A
19 = t
B
F b1
31 = b
F a1x
C
F a1y
20 = d
F a2y
D
Fa2x
2
Fc2x
F b2
C
D
Fd2
F c2y
FIGURE 5-4A
Dimensions and Free Body Diagram for Problem 5-4
1.
From Problem 4-4, the principal stresses in the shank of the ball where it joins the ball bracket are:
σ1  114  MPa
σ2  0  MPa
σ3  0  MPa
2. Since 1 is the only nonzero principal stress, it is also the von Mises stress.The factor of safety against a static
failure at the shank of the ball is
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
σ'  σ1
3.
Na 
5-4-2
Sy
Na  2.6
σ'
From Problem 4-4, the principal stresses at the bearing area in the ball bracket hole are:
σ1  9.93 MPa
σ2  0  MPa
σ3  0  MPa
4. Since 1 is the only nonzero principal stress, it is also the von Mises stress.The factor of safety against a static
bearing failure in the ball bracket hole is
σ'  σ1
Nb 
Sy
Nb  30.2
σ'
Tearout length
5. From Problem 4-4, the shear stress in the tearout area in
the ball bracket is:
τ  4.41 MPa
2
6. For pure shear, the von Mises stress is σ'  3  τ
and the factor of safety against a static tearout failure is
Nc 
Sy
Nc  39.3
σ'
7. From Problem 4-4, the principal stresses in the attachment
bolts if they are 19-mm diameter are:
σx  53.6 MPa
σy  0  MPa
R
d
FIGURE 5-4B
Tearout Diagram for Problem 5-4
τxy  1.7 MPa
8.
The von Mises stress and the factor of safety against a static failure in the attachment bolts are:
σ' 
2
σ'  53.7 MPa
9.
2
σx  σy  σx σy  3  τxy
2
Nd 
Sy
σ'
Nd  5.6
From Problem 4-4, the principal stresses in the bracket due to bending in the ball bracket as a cantilever are:
σ1  72.8 MPa
σ2  0  MPa
σ3  0  MPa
10. Since 1 is the only nonzero principal stress, it is also the von Mises stress.The factor of safety against a static
bearing failure in the ball bracket hole is
σ'  σ1
Ne 
Sy
σ'
Ne  4.1
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-5-1
PROBLEM 5-5
Statement:
Repeat Problem 5-4 for the loading conditions of Problem 3-5, i.e., determine the horizontal force
that will result on the ball from accelerating a 2000-kg trailer to 60 m/sec in 20 sec. Assume a
constant acceleration. From Problem 3-5, the pull force is 6000 N. Determine static safety factors
for:
(a) The shank of the ball where it joins the ball bracket.
(b) Bearing failure in the ball bracket hole.
(c) Tearout failure in the ball bracket.
(d) Tensile failure in the 19-mm diameter attachment holes.
(e) Bending failure in the ball bracket as a cantilever.
Given:
a  40 mm
b  31 mm
Mtongue  100  kg Fpull  6  kN
c  70 mm
d  20 mm
d sh  26 mm
t  19 mm
S y  300  MPa
Assumptions: 1. The nuts are just snug-tight (no pre-load), which is the worst case.
2. All reactions will be concentrated loads rather than distributed loads or pressures.
Solution:
See Figures 5-5 and Mathcad file P0505.
W tongue
70 = c
1
F pull
1
40 = a
2
A
B
A
19 = t
B
F b1
31 = b
F a1x
C
F a1y
20 = d
F a2y
D
Fa2x
2
Fc2x
F b2
C
D
Fd2
F c2y
FIGURE 5-5A
Dimensions and Free Body Diagram for Problem 5-5
1. From Problem 4-5, the principal stresses in the shank of the ball where it joins the ball bracket are:
σ1  139  MPa
σ2  0  MPa
σ3  0  MPa
2. Since 1 is the only nonzero principal stress, it is also the von Mises stress.The factor of safety against a static
failure at the shank of the ball is
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
σ'  σ1
Na 
5-5-2
Sy
Na  2.2
σ'
3. From Problem 4-5, the principal stresses at the bearing area in the ball bracket hole are:
σ1  12.15  MPa
σ2  0  MPa
σ3  0  MPa
4. Since 1 is the only nonzero principal stress, it is also the von Mises stress.The factor of safety against a static
bearing failure in the ball bracket hole is
σ'  σ1
Nb 
Sy
Nb  24.7
σ'
Tearout length
5. From Problem 4-5, the shear stress in the tearout area in
the ball bracket is:
τ  5.4 MPa
2
6. For pure shear, the von Mises stress is σ'  3  τ
and the factor of safety against a static tearout failure is
Nc 
Sy
Nc  32.1
σ'
7. From Problem 4-5, the principal stresses in the attachment
bolts if they are 19-mm diameter are:
σx  64.2 MPa
σy  0  MPa
R
d
FIGURE 5-5B
Tearout Diagram for Problem 5-5
τxy  1.7 MPa
8. The von Mises stress and the factor of safety against a static failure in the attachment bolts are:
σ' 
2
2
σx  σy  σx σy  3  τxy
σ'  64.3 MPa
2
Nd 
Sy
σ'
Nd  4.7
9. From Problem 4-5, the principal stresses in the bracket due to bending in the ball bracket as a cantilever are:
σ1  85.1 MPa
σ2  0  MPa
σ3  0  MPa
10. Since 1 is the only nonzero principal stress, it is also the von Mises stress.The factor of safety against a static
bearing failure in the ball bracket hole is
σ'  σ1
Ne 
Sy
σ'
Ne  3.5
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-6-1
PROBLEM 5-6
Statement:
Repeat Problem 5-4 for the loading conditions of Problem 3-6, i.e., determine the horizontal force
that will results from an impact between the ball and the tongue of the 2000-kg trailer if the hitch
deflects 2.8 mm dynamically on impact. The tractor weighs 1000 kg and the velocity at impact is
m/sec. Determine static safety factors for:
(a) The shank of the ball where it joins the ball bracket.
(b) Bearing failure in the ball bracket hole.
(c) Tearout failure in the ball bracket.
(d) Tensile failure in the 19-mm diameter attachment holes.
(e) Bending failure in the ball bracket as a cantilever.
Given:
a  40 mm
b  31 mm
c  70 mm
d  20 mm
Mtongue  100  kg Fpull  55.1 kN
d sh  26 mm
t  19 mm
S y  300  MPa
Assumptions: 1. The nuts are just snug-tight (no pre-load), which is the worst case.
2. All reactions will be concentrated loads rather than distributed loads or pressures.
Solution:
See Figures 5-6 and Mathcad file P0506.
W tongue
70 = c
1
F pull
1
40 = a
2
A
B
A
19 = t
B
F b1
31 = b
F a1x
C
F a1y
20 = d
F a2y
D
Fa2x
2
Fc2x
F b2
C
D
Fd2
F c2y
FIGURE 5-6A
Dimensions and Free Body Diagram for Problem 5-6
1. From Problem 4-6, the principal stresses in the shank of the ball where it joins the ball bracket are:
σ1  1277 MPa
σ2  0  MPa
σ3  0  MPa
2. Since 1 is the only nonzero principal stress, it is also the von Mises stress.The factor of safety against a static
failure at the shank of the ball is
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
σ'  σ1
Na 
5-6-2
Sy
Na  0.23
σ'
3. From Problem 4-6, the principal stresses at the bearing area in the ball bracket hole are:
σ1  111.5  MPa
σ2  0  MPa
σ3  0  MPa
4. Since 1 is the only nonzero principal stress, it is also the von Mises stress.The factor of safety against a static
bearing failure in the ball bracket hole is
σ'  σ1
Nb 
Sy
Nb  2.7
σ'
Tearout length
5. From Problem 4-6, the shear stress in the tearout area in
the ball bracket is:
τ  49.6 MPa
2
6. For pure shear, the von Mises stress is σ'  3  τ
and the factor of safety against a static tearout failure is
Nc 
Sy
Nc  3.5
σ'
7. From Problem 4-6, the principal stresses in the attachment
bolts if they are 19-mm diameter are:
σx  540.5  MPa
σy  0  MPa
R
d
FIGURE 5-6B
Tearout Diagram for Problem 5-6
τxy  1.7 MPa
8. The von Mises stress and the factor of safety against a static failure in the attachment bolts are:
σ' 
2
2
σx  σy  σx σy  3  τxy
σ'  540.5 MPa
2
Nd 
Sy
σ'
Nd  0.56
9. From Problem 4-6, the principal stresses in the bracket due to bending in the ball bracket as a cantilever are:
σ1  635.5  MPa
σ2  0  MPa
σ3  0  MPa
10. Since 1 is the only nonzero principal stress, it is also the von Mises stress.The factor of safety against a static
bearing failure in the ball bracket hole is
σ'  σ1
Ne 
Sy
σ'
Ne  0.47
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-7-1
PROBLEM 5-7
Statement:
Design the wrist pin of Problem 3-7 for a safety factor of 3 and S y = 100 ksi if the pin is hollow and
loaded in double shear.
Given:
Force on wrist pin
Fwristpin  12.258 kN
Yield strength
S y  100  ksi
Design safety factor
Nd  3
od  0.375  in
Assumptions: Choose a suitable outside diameter, say
Solution:
Fwristpin  2756 lbf
See Figure 4-12 in the text and Mathcad file P0507.
1. The force at each shear plane is
F 
Fwristpin
F  1378 lbf
2
2. With only the direct shear acting on the plane, the Mohr diagram will be a circle with center at the origin and
radius equal to the shear stress. Thus, the principal normal stress is numerically equal to the shear stress, which in
this case is also the principal shear stress, so we have  = 1 = '.
3. The shear stress at each shear plane is
4. Using the distortion-energy failure theory,
5. Solving for the inside diameter,
F
τ=
id 
=
A
4 F

2
π od  id
Sy
Nd =
σ'
2
od 

= σ'

2
2

2
π od  id  S y
=
4 F
4  F  Nd
π S y
6. Round this down to the decimal equivalent of a common fraction (9/32),
7. The realized factor of safety is,
N 

2

id  0.281  in
2
π od  id  S y
4 F
id  0.297 in
N  3.5
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-8-1
PROBLEM 5-8
Statement:
Given:
A paper mill processes rolls of paper having a density of 984 kg/m3. The paper roll is 1.50-m
outside diameter (OD) x 0.22-m inside diameter (ID) x 3.23-m long and is on a simple supported,
hollow, steel shaft with S y = 300 MPa. Find the shaft ID needed to obtain a static safety factor of 5
if the shaft OD is 22 cm.
Paper roll:
Density
ρ  984 
kg
y
3
Outside dia.
m
OD  1500 mm
Inside dia.
ID  220  mm
Length
L  3230 mm
w
x
L
R
R
Shaft:
Strength
V
S y  300  MPa
Outside dia.
od  220  mm
Factor of safety
Ns  5
R
L/2
x
-R
Assumptions: 1. The shaft is stiffer than the paper roll so the
weight of the roll on the shaft can be modelled as
a uniformly distributed load.
2. The bearings that support the shaft are close to
the ends of the paper roll and are thin with
respect to the length of the roll so we can
consider the distance between the shaft supports
to be the same as the length of the roll.
Solution:
L
0
M
2
wL /8
x
0
L/2
See Figure 5-8 and Mathcad file P0508.
L
FIGURE 5-8
Load, Shear, and Moment Diagrams
for Problem 5-8
1. The weight of the paper roll is,

4
π
2
Volume
V 
Weight
W  ρ  g  V
2

3
 OD  ID  L
V  5.585  m
(1)
W  53.895 kN
(2)
2. From Figure 5-8, we see that the bending moment in the shaft is a maximum at the center of the span. First,
determine the magnitude of the distributed load, then find the maximum bending moment using Figure D-2(b) in
Appendix B with a = 0 and x = L/2.
Distributed load
w 
W
w  16.686
L
2
Maximum moment
Mmax 
w L
newton
7
Mmax  2.176  10  newton  mm
8
(3)
mm
(4)
3. Using equation 4.11b, find the maximum bending stress as a function of the unkown shaft inside diameter, id.
Bending stress
at midspan
σmax =
M c
I
=
32 Mmax od

4

(5)
4
π od  id
4. This is the only stress element present at this point on the shaft and there is no shear stress at this point so max
= 1 and 2 = 3 = 0. Furthermore, since 2 and 3 are zero, max = '. Equation 5.8a can be used to find the
unknown id,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
Factor of safety
Ns =
5-8-2
Sy
(6)
σ'
Substituting equation 5 into 6 and solving for id, we have
1
Shaft id
 π Sy od4  32 Ns Mmax od 
id 


π S y


4
id  198  mm
(7)
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-9-1
PROBLEM 5-9
Statement:
A ViseGrip plier-wrench is drawn to scale in Figure P5-3, and for which the forces were analyzed in
Problem 3-9 and the stresses in Problem 4-9, find the safety factors for each pin for an assumed
clamping force of P = 4000 N in the position shown. The pins are 8-mm dia, S y = 400 MPa, and are
all in double shear.
Given:
Pin stresses as calculated in Problem 4-9:
Pin 1-2
τ12  74.6 MPa
Pin 1-4
τ14  50.7 MPa
Pin 2-3
τ23  50.7 MPa
Pin 3-4
τ34  50.7 MPa
S y  400  MPa
Yield strength
Assumptions: Links 3 and 4 are in a toggle position, i.e., the pin that joins links 3 and 4 is in line with the pins th
join 1 with 4 and 2 with 3.
Solution:
1.
See Figure 5-9 and Mathcad file P0509.
The FBDs of the assembly and each individual link are shown in Figure 5-9. The dimensions, as scaled from
Figure P5-3 in the text, are shown on the link FBDs.
4
F
P
1
2
3
P
F
55.0 = b
50.0 = a
39.5 = c
F
F14
22.0 = d
129.2°
1

4
F34
F41
F21
P

28.0 = e


F43

F12
3
F23
F32
P
2.8 = g
21.2 = h
2
F
26.9 = f
FIGURE 5-9
Free Body Diagrams for Problem 5-9
2.
The pins are in pure shear, so the principal stresses are
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3.
5-9-2
Pin joining 1 and 2
σ'12 
3  τ12
σ'12  129.211 MPa
All other pins
σ'14 
3  τ14
σ'14  87.815 MPa
Using the distortion-energy failure theory, the factors of safety are
Pin joining 1 and 2
All other pins
N12 
N14 
Sy
σ'12
Sy
σ'14
N12  3.1
N14  4.6
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-10-1
PROBLEM 5-10
Statement:
Given:
An over-hung diving board is shown in Figure P5-4a. Assume cross-section dimensions of 305 mm
x 32 mm. Find the largest principal stress in the board when a 100-kg person is standing at the free
end. What is the static safety factor if the material is brittle fiberglass with S ut = 130 MPa in the
longitudinal direction?
Maximum principal stresses due to
bending at R2 from Problem 4-10
2000 = L
R1
P
σ1  24.5 MPa
σ2  0  MPa
R2
σ3  0  MPa
Ultimate strength
S ut  130  MPa
700 = a
FIGURE 5-10
Free Body Diagram for Problem 5-10
Solution:
1.
See Figure 5-10 and Mathcad file P0510.
The diving board will be in tension at the top of the board and compression along the bottom. At the top, over
the right-hand support, the nonzero principal stress is positive and the load line on the 1-3 diagram is along
the 1 axis. Using the Modified-Mohr failure theory, the static safety factor is
Ns 
S ut
σ1
Ns  5.3
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-11-1
PROBLEM 5-11
Statement:
Given:
Repeat Problem 5-10 assuming the 100-kg person in Problem 5-10 jumps up 25 cm and lands back
on the board. Assume the board weighs 29 kg and deflects 13.1 cm statically when the person
stands on it. What is the static safety factor if the material is brittle fiberglass with S ut = 130 MPa i
the longitudinal direction?
Maximum principal stresses due to
bending at R2 from Problem 4-11
2000 = L
R1
P
σ1  76.3 MPa
σ2  0  MPa
R2
σ3  0  MPa
Ultimate strength
700 = a
S ut  130  MPa
FIGURE 5-11
Free Body Diagram for Problem 5-11
Solution:
1.
See Figure 5-11 and Mathcad file P0511.
The diving board will be in tension at the top of the board and compression along the bottom. At the top, ove
the right-hand support, the nonzero principal stress is positive and the load line on the 1-3 diagram is along
the 1 axis. Using the Modified-Mohr failure theory, the static safety factor is
Ns 
S ut
σ1
Ns  1.7
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-12-1
PROBLEM 5-12
Statement:
Given:
Repeat Problem 5-10 using the cantilevered diving board design in Figure P5-4b.
2000
Maximum principal stresses due to
bending at support from Problem 4-12
1300 = L
σ1  24.5 MPa
P
σ2  0  MPa
M1
σ3  0  MPa
Ultimate strength
Solution:
S ut  130  MPa
See Figure 5-12 and Mathcad file P0512.
R1
700
FIGURE 5-12
Free Body Diagram for Problem 5-12
1.
The diving board will be in tension at the top of the board and compression along the bottom. At the top, at
the built-in support, the nonzero principal stress is positive and the load line on the 1-3 diagram is along the
1 axis. Using the Modified-Mohr failure theory, the static safety factor is
Ns 
S ut
σ1
Ns  5.3
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-13-1
PROBLEM 5-13
Statement:
Given:
Repeat Problem 5-11 using the cantilevered diving board design in Figure P5-4b. Assume the
board weighs 19 kg and deflects 8.5 cm statically when the person stands on it.
2000
Maximum principal stresses due to
bending at support from Problem 4-13
1300 = L
σ1  87.1 MPa
P
σ2  0  MPa
M1
σ3  0  MPa
Ultimate strength
Solution:
S ut  130  MPa
See Figure 5-13 and Mathcad file P0513.
R1
700
FIGURE 5-13
Free Body Diagram for Problem 5-13
1.
The diving board will be in tension at the top of the board and compression along the bottom. At the top, at the
built-in support, the nonzero principal stress is positive and the load line on the 1-3 diagram is along the 1
axis. Using the Modified-Mohr failure theory, the static safety factor is
Ns 
S ut
σ1
Ns  1.5
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-14-1
PROBLEM 5-14
Statement:
Figure P4-5 shows a child's toy called a pogo stick. The child stands on the pads, applying half he
weight on each side. She jumps off the ground, holding the pads up against her feet, and bounces
along with the spring cushioning the impact and storing energy to help each rebound. Design the
aluminum cantilever beam sections on which she stands to survive jumping 2 in off the ground
with a safety factor of 2. Use 1100 series aluminum. Define and size the beam shape.
Given:
Cold rolled 1100 aluminum:
Yield strength
S y  22 ksi
Safety factor
Ns  2
Assumptions: The beam will have a rectangular
cross-section with the load applied at a
distance of 5 in from the central support.
L  5  in
Solution:
See Figure 5-14 and Mathcad file P0514.
1. From Problem 3-14, the total dynamic force on both
foot supports is
Fi /2
Fi /2
Fi  224  lbf
Therefore, the load on each support is
P 
Fi
P  112  lbf
2
2. To give adequate support to the childs foot, let the
width of the support beam be
w  1.5 in
3. From Figure B-1(a) in Appendix B, the maximum
bending moment at x = 0 is
M  P L
P
FIGURE 5-14
M  560  in lbf
Free Body Diagram for Problem 5-14
4. We can now calculate the minimum required section modulus, Z = I/c. Using the distortion-energy failure theor
the bending stress will also be the only nonzero principal stress, which will also be the von Mises stress.
Design equation
Ns =
Bending stress
σ=
Solving for Z,
5. For a rectangular cross-section,
Solving for t,
Z 
Sy
σ'
M
Z
Sy
Ns
N s M
I=
t 
= σ' =
Z  834.3  mm
Sy
w t
3
12
6 Z
w
and
c=
t
2
so
Z=
3
w t
2
6
t  0.451  in
Round this up to the next higher decimal equivalent of a common fraction,
t  0.500  in
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-15-1
PROBLEM 5-15
Statement:
What is the safety factor for the shear pin as defined in Problem 4-15?
Solution:
Any part whose stress equals its strength has a safety factor of 1 by definition.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-16-1
PROBLEM 5-16
Statement:
A track to guide bowling balls is designed with two round rods as shown in Figure P5-6. The rods
are not parallel to one another but have a small angle between them. The balls roll on the rods unt
they fall between them and drop onto another track. The angle between the rods is varied to cause
the ball to drop at different locations. Find the static safety factor for the 1-in dia SAE 1045
normalized steel rods.
(a) Assume rods are simply supported at each end.
(b) Assume rods are fixed at each end.
S y  58 ksi
Given:
Yield strength
Solution:
See Figure 5-16 and Mathcad file P0516.
Fball
a
R1
1. The maximum bending stress will occur at the outer
fibers of the rod at the section where the maximum
bending moment occurs which, in this case, is at x = a.
The only stress present on the top or bottom surface of
the rod is the bending stress x. Therefore, on the
R2
L
FIGURE 5-16A
Free Body Diagram for Problem 5-16(a), taken
on a plane through the rod axis and ball center
bottom surface where the stress is tensile, x is the
principal stress 1 . Thus, from Problem 4-16, for a simply
supported rod,
Maximum principal stress
σ1  748  psi
σ'a  σ1
2. Using the distortion-energy failure theory, the safety factor against a static failure is
Nsa 
Sy
σ'a
Nsa  78
3. For the built-in case, the maximum bending stress
will occur at the outer fibers of the rod at the section
where the maximum bending moment occurs which, in
this case, is at x = L. The only stress present on the
top or bottom surface of the rod is the bending stress
x. Therefore, on the bottom surface where the stress
is tensile, x is the principal stress 1 . Thus, from
Problem 4-16, for a simply supported rod,
Maximum principal stress
Fball
a
M1
R1
L
R 2 M2
FIGURE 5-16B
Free Body Diagram for Problem 5-16(b), taken on a
plane through the rod axis and ball center
σ1  577  psi
σ'b  σ1
4. Using the distortion-energy failure theory, the safety factor against a static failure is
Nsb 
Sy
σ'b
Nsb  101
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-17-1
PROBLEM 5-17
Statement:
Given:
A pair of ice tongs is shown in Figure P5-7. The ice weighs 50 lb and is 10 in wide across the
tongs. The distance between the handles is 4 in, and the mean radius r of the tong is 6 in. The
rectangular cross-sectional dimensions are 0.75 x 0.312 in. Find the safety factor for the tongs if
their S y = 30 ksi.
F
Yield strength
C
S y  30 ksi
FC
O
See Problem 4-17, Figure 5-17, and
Solution:
Mathcad file P0517.
11.0 = ax
1. The maximum bending stress in the tong was found
in Problem 4-17 at point A.
Vertical direction
3.5 = cy
FO
2.0 = cx
A
12.0 = by
σi  8.58 ksi
5.0 = bx
FB
All other components are zero
B
2. There are no other stress components present so
σ1  σi
σ2  0  ksi
σ3  0  ksi
and
σ'  σ1
σ'  8.58 ksi
W/2
FIGURE 5-17
3. The factor of safety is (using the distortion energy theory)
Free Body Diagram for Problem 5-17
N 
Sy
σ'
N  3.5
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-18-1
PROBLEM 5-18
Statement:
A pair of ice tongs is shown in Figure P5-7. The ice weighs 50 lb and is 10 in wide across the
tongs. The distance between the handles is 4 in, and the mean radius r of the tong is 6 in. The
rectangular cross-sectional dimensions are 0.75 x 0.312 in. Find the safety factor for the tongs if
they are made from Class 20 gray cast iron.
F
Given:
Tensile strength
S ut  22 ksi
Compressive strength
S uc  83 ksi
C
FC
O
See Problem 4-18, Figure 5-18, and
Solution:
Mathcad file P0518.
1. The maximum bending stress in the tong was found
in Problem 4-17 at point A.
All other components are zero
12.0 = by
5.0 = bx
FB
B
2. Therefore, the principal stresses are
σ2  0  ksi
2.0 = cx
A
σi  8.58 ksi
Vertical direction
σ1  σi
11.0 = ax
3.5 = cy
FO
W/2
σ3  0  ksi
3. The load line on the 1-3 diagram is along the 1 axis.
Using the Modified-Mohr failure theory, the static safety
factor is
S ut
N 
N  2.6
FIGURE 5-18
Free Body Diagram for Problem 5-18
σ1
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-19-1
PROBLEM 5-19
Statement:
Determine the size of the clevis pin, shown in Figure P5-8, needed to withstand an applied force
of 130 000 lb. Also determine the required outside radius of the clevis end to not fail in either tear
out or bearing if the clevis flanges are each 2.5 in thick. Use a safety factor of 3 for all modes of
failure. Assume S y = 89.3 ksi for the pin and S y = 35.5 ksi for the clevis.
Given:
Applied force
Clevis strength
P  130  kip
S yclevis  35.5 ksi
Safety factor
Ns  3
Solution:
1.
Clevis flange thickness
t  2.50 in
Pin strength
S ypin  89.3 ksi
See Figures P5-8 in the text and Mathcad file P0519.
Determine the force carried by each of the two flanges of the clevis.
F  0.5 P
F  65 kip
This force is transmitted through each end of the clevis pin, which is in double shear.
2.
σ'pin = 3  τpin =
The pin is in direct (pure) shear. Therefore, the von Mises stress is
4 3 F
π d
3.
Calculate the minimum required clevis pin diameter using the distortion-energy failure theory.
Ns =
S ypin
σ'pin
2
=
Solving for the pin diameter
π d  S ypin
4 3 F
4  3  F  Ns
d 
d  2.194 in
π S ypin
Round this up to the next higher decimal equivalent of a common fraction ( 2 1/4)
4.
2
d  2.250  in
Check the bearing stress in the clevis due to the pin on one side of the clevis.
2
Bearing stress area
Ab  d  t
Ab  5.625 in
Bearing force
Fb  F
Fb  65 kip
Bearing stress
σb 
Fb
σb  11.6 ksi
Ab
Tearout length
5. Determine the safety factor against a static bearing failure.
Nbear 
S yclevis
σb
Nbear  3.1
Since this is greater than 3, the pin diameter is acceptable.
6. Determine the tearout stress in the clevis.
Shear area (see Figure 5-19)
2
Atear = 2  t R  ( 0.5 d )
2
d
R
FIGURE 5-19
Shear force
Ftear  F
Tearout Diagram for Problem 5-19
Ftear  65 kip
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
Shear stress and distortion-energy equation
Ns =
S yclevis
σ'tear
=
S yclevis
3  τtear
τtear =
2
=
5-19-2
2  t S yclevis R  ( 0.5 d )
Ftear
Atear
Ftear
=
2
2  t R  ( 0.5 d )
2
2
3  Ftear
2
Solving for the clevis radius, R
 3  Ftear Ns 
2
R  
  ( 0.5 d )
2

t
S

yclevis 

R  2.211 in
Round this up to the next higher decimal equivalent of a common fraction ( 2 1/4)
The tearout area for each flange is
2
Atear  2  t R  ( 0.5 d )
2
R  2.250  in
2
Atear  9.743 in
7. Design summary:
Pin diameter
d  2.250 in
Clevis flange radius
R  2.25 in
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-20-1
PROBLEM 5-20
Statement:
A 100 N-m torque is applied to a 1-m-long, solid, round shaft. Design it to limit its angular
deflection to 2 deg and select a steel alloy to have a yielding safety factor of 2.
Given:
Applied torque
Maximum deflection
T  100  N  m
θmax  2  deg
Safety factor
Ns  2
L  1000 mm
G  79 GPa
Shaft length
Modulus of rigidity
Assumptions: A ductile steel will be chosen.
Solution:
1.
See Mathcad file P0520.
Using the angular deflection requirement and equation (4.24), determine the required polar moment of inertia an
the minimum diameter.
θ=
TL
J 
J G
TL
4
J  3.626  10  mm
θmax G
4
1
J =
π d
d 
32
d  24.653 mm
d

2
τmax 
τmax  34.47  MPa
J
For this case of pure shear, use the distortion-energy theory and equations (5.8) and (5.9) to solve for the
minimum required yield strength.
Ns =
4.
4
Determine the shear stress at the outside diameter of the shaft using equation (4.23b).
T  
3.
 32 J 
 π 


d  25 mm
Round this up to
2.
4
Sy
σ'
=
Sy
3  τmax
S y 
3  τmax Ns
S y  119.4  MPa
Using this value of S y, choose a steel from Table A-9 in Appendix A.
Any of the steels listed in Table A-9 will be adequate. The least expensive is AISI 1010, hot rolled.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-21-1
PROBLEM 5-21
Statement:
Figure P5-9 shows an automobile wheel with two common styles of lug wrench being used to
tighten the wheel nuts, a single-ended wrench in (a), and a double-ended wrench in (b). The
distance between points A and B is 1 ft in both cases and the handle diameter is 0.625 in. What is
the maximum force possible before yielding the handle if the material S y = 45 ksi?
Given:
Distance between A and B
d AB  1  ft
Wrench diameter
Yield strength
d  0.625  in
S y  45 ksi
Assumptions: 1. The forces exerted by the user's hands lie in a plane through the wrench that is also parallel to
the plane of the wheel.
2. The applied torque is perpendicular to the plane of the forces.
3. By virtue of 1 and 2 above, this is a planar problem that can be described in a 2D FBD.
Solution:
See Figure 5-21 and Mathcad file P0521.
12" = dAB
1. From examination of the FBDs, we see that, in
both cases, the arms are in bending and the stub
that holds the socket wrench is in pure torsion.
The maximum bending stress in the arm will occur
near the point where the arm transitions to the
stub. The stress state at this transition is very
complicated, but we can find the nominal bending
stress there by treating the arm as a cantilever
beam, fixed at the transition point. For both cases
the torque in the stub is the same.
F
T
F
(a) Single-ended Wrench
12" = dAB
Case (a)
F
6"
2. The bending moment at the transition is
Ma = Fa d AB
T
3. The tensile stress at this point is found from
F
(b) Double-ended Wrench
Moment of inertia
I 
π d
4
FIGURE 5-21
4
I  0.00749 in
64
Free Body Diagrams for Problem 5-21
Dist to extreme fibre
c  0.5 d
Stress
σx =
c  0.313 in
Ma c
I
4. There are no other stress components present at this point, so x is the maximum principle stress here and
σ2  0  psi
σ1 = σx
σ3  0  psi
5. Since there is only one nonzero principal stress, the von Mises stress is
σ' = σ1 = σx =
Ma c
I
=
Fa d AB c
I
6. Using the distortion-energy theory, solve for the maximum applied force.
Ns =
Sy
σ'
=
I Sy
Fa d AB c
=1
Fa 
I  Sy
d AB c
Fa  89.882 lbf
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-21-2
σ' 
7. The von Mises stress in the handle at the transition point is
T  Fa d AB
8. Determine the torque in the stub.
Fa d AB c
σ'  45 ksi
I
T  1079 in lbf
9. The shear stress at any point on the outside surface of the stub is found from
Polar moment of inertia
J  2  I
Shear stress
τxy 
4
J  0.0150 in
Tc
τxy  22.5 ksi
J
10. There are no other stress components present along the outside surface of the stub, so
σ1  τxy
and
σ' 
σ1  22.5 ksi
2
σ1  σ1 σ3  σ3
σ2  0  psi
2
σ3  σ1
σ'  39.0 ksi
11. Thus, the maximum von Mises stress for case (a) is on the upper surface of the handle (arm) near the point
where it transitions to the stub, and the maximum force that can be applied to the handle without yielding is
Fa  89.9 lbf
Case (b)
12. The bending moment at the transition is
Mb =
11. The tensile stress at this point is found from
σx =
Fb d AB
2
Mb c
I
12. There are no other stress components present at this point, so x is the maximum principle stress here and
σ2  0  psi
σ1 = σx
σ3  0  psi
13. Since there is only one nonzero principal stress, the von Mises stress is σ' = σ1 = σx =
Mb c
I
=
Fb d AB c
2 I
14. Using the distortion-energy theory, solve for the maximum applied force.
Ns =
Sy
σ'
=
2 I  Sy
Fb d AB c
Fb 
=1
2 I Sy
Fb  179.763 lbf
d AB c
15. The von Mises stress in the handle at the transition point is
T  Fb d AB
16. The torque in the stub is
σ' 
Fb d AB c
2 I
σ'  45 ksi
T  2157 in lbf
14. The shear stress at any point on the outside surface of the stub is found from
Shear stress
τxy 
Tc
J
τxy  45 ksi
15. There are no other stress components present along the outside surface of the stub, so
σ1  τxy
σ1  45.0 ksi
σ2  0  psi
σ3  σ1
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
and
σ' 
2
σ1  σ1 σ3  σ3
5-21-3
2
σ'  77.9 ksi
16. Since the von Mises stress in the stub due to torsion is greater than the yield strength, the force in the
handle will be limited by the shear stress in the stib and by the bending stress in the handle.
Ns =
Fb 
Sy
σ'
=
Sy
3  τxy
J  Sy
3  d AB c
=
J  Sy
3 T  c
=
J  Sy
3  Fb d AB c
=1
Fb  103.8 lbf
17. Thus, the maximum von Mises stress for case (b) is on the stub, and the maximum force that can be
applied to the handles without yielding is
Fb  103.8 lbf
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-22-1
PROBLEM 5-22
Statement:
A roller-blade skate is shown in Figure P5-10. The polyurethane wheels are 72 mm dia and
spaced on 104-mm centers. The skate-boot-foot combination weighs 2 kg. The effective "spring
rate" of the person-skate subsystem is 6000 N/m. The axles are 10-mm-dia steel pins in double
shear with S y = 400 MPa. Find the safety factor for the pins when a 100-kg person lands a 0.5-m
jump on one foot.
(a) Assume all 4 wheels land simultaneously.
(b) Assume that one wheel absorbs all the landing force.
Given:
Axle pin diameter
Solution:
See Figure P5-10 and Mathcad file P0522.
d  10 mm
1.
From Problem 4-22, we have the stresses for cases (a) and (b):
2.
Using the distortion-energy failure theory,
Case (a) all wheels landing
Nsa 
Case (b) one wheel landing
Nsb 
Sy
3  τa
Sy
3  τb
Yield strength
S y  400  MPa
τa  5.71 MPa
τb  22.9 MPa
Nsa  40.4
Nsb  10.1
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-23a-1
PROBLEM 5-23a
Statement:
Given:
A beam is supported and loaded as shown in Figure P5-11a. For the data given in row a from
Table P5-2, find the static safety factor:
(a) If the beam is a ductile material with S y = 300 MPa,
(b) If the beam is a cast-brittle material with S ut = 150 MPa, S uc = 570 MPa.
Ductile yield strength
L
S y  300  MPa
b
Brittle ultimate tensile strength
S ut  150  MPa
Solution:
a
F
w
See Figure 5-23 and Mathcad file P0523a.
R2
R1
FIGURE 5-23
Free Body Diagram for Problem 5-23
1.
The maximum bending stress occurs under the concentrated load F at x = b. It was determined in Problem 4-23a
as
σx  88.7 MPa
2.
Since this is the only stress component present in the given coordinate frame, x is equal to 1 and the other two
principal stresses are zero.
σ1  σx
3.
4.
σ2  0  MPa
σ3  0  MPa
For case (a), use the distortion-energy failure theory. With only one nonzero principal stress, the von Mises
stress is the same as 1.
von Mises stress
σ'  σ1
Safety factor, case (a)
Nsa 
Sy
σ'
σ'  88.7 MPa
Nsa  3.4
For case (b), use the Modified Mohr failure theory. The nonzero principal stress is positive and the load line
on the s1-s3 diagram is along the 1 axis.
Safety factor, case (b)
Nsb 
S ut
σ1
Nsb  1.7
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-24a-1
PROBLEM 5-24a
Statement:
A beam is supported and loaded as shown in Figure P5-11b. For the data given in row a from Tabl
P5-2, find the static safety factor:
(a) If the beam is a ductile material with S y = 300 MPa,
(b) If the beam is a cast-brittle material with S ut = 150 MPa, S uc = 570 MPa.
L
Given:
Ductile yield strength
a
S y  300  MPa
F
Brittle ultimate strength
S ut  150  MPa
w
M1
Solution:
See Figure 5-24 and Mathcad file P0524a.
R1
FIGURE 5-24
Free Body Diagram for Problem 5-24
1.
The maximum bending stress occurs at the support where x = 0. It was determined in Problem 4-24a as
σx  410  MPa
2.
Since this is the only stress component present in the given coordinate frame, x is equal to 1 and the other two
principal stresses are zero.
σ1  σx
3.
4.
σ2  0  MPa
σ3  0  MPa
For case (a), use the distortion-energy failure theory. With only one nonzero principal stress, the von Mises
stress is the same as 1.
von Mises stress
σ'  σ1
Safety factor, case (a)
Nsa 
Sy
σ'
σ'  410 MPa
Nsa  0.73
For case (b), use the Modified Mohr failure theory. The nonzero principal stress is positive and the load line on
the 1-3 diagram is along the 1 axis.
Safety factor, case (b)
Nsb 
S ut
σ1
Nsb  0.37
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-25a-1
PROBLEM 5-25a
Statement:
A beam is supported and loaded as shown in Figure P5-11c. For the data given in row a from
Table P5-2, find the static safety factor:
(a) If the beam is a ductile material with S y = 300 MPa,
(b) If the beam is a cast-brittle material with S ut = 150 MPa, S uc = 570 MPa.
L
Given:
Ductile yield strength
S y  300  MPa
b
Brittle ultimate strength
S ut  150  MPa
Solution:
a
F
w
See Figure 5-25 and Mathcad file P0525a.
R2
R1
FIGURE 5-25
Free Body Diagram for Problem 5-25
1.
The maximum bending stress occurs at the right-hand support where x = b. It was determined in Problem 4-25a
as
σx  151.6  MPa
2.
Since this is the only stress component present in the given coordinate frame, x is equal to 1 and the other
two principal stresses are zero.
σ1  σx
3.
4.
σ2  0  MPa
σ3  0  MPa
For case (a), use the distortion-energy failure theory. With only one nonzero principal stress, the von Mises
stress is the same as 1.
von Mises stress
σ'  σ1
Safety factor, case (a)
Nsa 
Sy
σ'
σ'  151.6 MPa
Nsa  2.0
For case (b), use the Modified Mohr failure theory. The nonzero principal stress is positive and the load line on
the 1-3 diagram is along the 1 axis.
Safety factor, case (b)
Nsb 
S ut
σ1
Nsb  0.99
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-26a-1
PROBLEM 5-26a
Statement:
A beam is supported and loaded as shown in Figure P5-11d. For the data given in row a from Tabl
P5-2, find the static safety factor:
(a) If the beam is a ductile material with S y = 300 MPa,
(b) If the beam is a cast-brittle material with S ut = 150 MPa, S uc = 570 MPa.
L
Given:
Ductile yield strength
S y  300  MPa
b
Brittle ultimate strength
S ut  150  MPa
Solution:
1.
a
F
w
See Figure 5-26 and Mathcad file P0526a.
R2
R1
R3
FIGURE 5-26
The maximum bending stress occurs under the concentrated load F, where x = a. It was determined in
Problem 4-26a as
Free Body Diagram for Problem 5-26
σx  31.5 MPa
2.
Since this is the only stress component present in the given coordinate frame, x is equal to 1 and the other two
principal stresses are zero.
σ1  σx
3.
4.
σ2  0  MPa
σ3  0  MPa
For case (a), use the distortion-energy failure theory. With only one nonzero principal stress, the von Mises
stress is the same as 1.
von Mises stress
σ'  σ1
Safety factor, case (a)
Nsa 
Sy
σ'
σ'  31.5 MPa
Nsa  9.5
For case (b), use the Modified Mohr failure theory. The nonzero principal stress is positive and the load line on
the 1-3 diagram is along the 1 axis.
Safety factor, case (b)
Nsb 
S ut
σ1
Nsb  4.8
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-27-1
PROBLEM 5-27
Statement:
A storage rack is to be designed to hold the paper roll of Problem 5-8 as shown in Figure P5-12.
Determine suitable values for dimensions a and b in the figure. Make the static factor of safety at
least 1.5. The mandrel is solid and inserts halfway into the paper roll.
(a) The beam is a ductile material with Sy = 300 MPa
(b) The beam is a cast-brittle material with Sut = 150 MPa, S uc = 570 MPa.
Given:
Paper roll dimensions
Ductile yield strength
3
OD  1.50 m
ID  0.22 m
Lroll  3.23 m
Factor of safety
Ns  1.5
S y  300  MPa
Brittle ultimate strength
S ut  150  MPa
Roll density
Assumptions: The paper roll's weight creates a
concentrated load acting at the tip of the
y
mandrel. The mandrel's root in the
w
stanchion experiences a distributed load
a
over the length of engagement (see the
solution to Problem 3-27 for further
discussion of this point). The required
diameter a of the mandrel root section
b
(over the length b) will be sized to use
the allowable tensile strength in bending.
R
The length b will be sized to use the
FIGURE 5-27
allowable transverse shear strength.
ρ  984  kg m
W
x
Lm
Free Body Diagram used in Problem 5-27
Solution:
1.
2.
See Figure 5-27 and Mathcad file P0527.
Determine the weight of the roll and the length of the mandrel.
Length
Lm  0.5 Lroll
2  W  Lm
b

W  53.9 kN
Lm  1.615 m
Mmax  W  Lm
Mmax  87.04 kN  m
Part (a) - The bending stress will be a maximum at the top or bottom of the mandrel at a section where the
mandrel root leaves the stanchion.
Mmax a
where
2 I
I=
π a
4
so,
64
σmax =
32 Mmax
π a
3
At this point the only nonzero stress component is max therefore
σ2  0  MPa
σ1 = σmax
5.
2
 OD  ID  Lroll  ρ  g
The maximum internal shear and moment occur at a section where the mandrel root leaves the stanchion. and
are
σmax =
4.
2
W 
Vmax =
3.

4
π
Weight
σ3  0  MPa
All three of the ductile failure theories have the same fail/safe boundary for this condition (slope of load line is
zero)
Ns =
Sy
σ1
or
Ns σ1 = S y
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-27-2
1
6.
7.
Solving for a,
 32 Ns W  Lm 
a  

 π S y 
Round this to
a  166  mm
4  Vmax
3 A
8  W  Lm
=
 π a 2 
 b
 4 
3 
At this point, this is the only nonzero stress component therefore, the principal stresses are
σ2  0  MPa
σ1 = τmax
9
a  164.272 mm
Using this value of a and equation (4.15c), solve for the shear stress on the neutral axis at the same section.
τmax =
8.
3
σ3 = τmax
Using the distortion energy theory, the von Mises stress is
σ' = 3  τmax
and
b 
Solving for b
Ns =
Sy
σ'
8  Ns W  Lm
Sy
3
b  92.9 mm
 π a 2  S y

3 
 4  3
a  166 mm
Rounding to higher even values, let
Ns τmax =
or
b  94 mm
for case (a).
10. Part (b) - The bending stress will be a maximum at the top or bottom of the mandrel at a section where the
mandrel root leaves the stanchion.
σmax =
Mmax a
where
2 I
I=
π a
4
so,
64
σmax =
32 Mmax
π a
3
11. At this point the only nonzero stress component is max therefore
σ2  0  MPa
σ1 = σmax
σ3  0  MPa
12. All three of the brittle failure theories have the same fail/safe boundary for this condition (slope of load line is
zero)
Ns =
S ut
σ1
=
S ut
σmax
=
2  I  S ut
Mmax a
3
=
π a  S ut
32 Mmax
1
13. Solving for a,
Round this to
 32 Ns Mmax 
a  

 π Sut 
3
a  206.97 mm
a  208  mm
14. Using this value of a and equation (4.15c), solve for the shear stress on the neutral axis at the same section.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
τmax =
4  Vmax
3 A
5-27-3
8  W  Lm
=
 π a 2 
 b
3 
 4 
15. At this point, this is the only nonzero stress component therefore, the principal stresses are
σ2  0  MPa
σ1 = τmax
σ3 = τmax
16. Using the Modified Mohr theory,
Ns =
Solving for b
S ut
σ1
=
 π a 2 
  b S ut
 4 
3 
S ut
τmax
=
b 
8  W  Lm
8  Ns W  Lm
 π a 2 
  Sut
3 
 4 
Rounding to higher even values, let
a  208 mm
b  68.3 mm
b  70 mm
for case (b).
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-28-1
PROBLEM 5-28
Statement:
Figure P5-13 shows a forklift truck negotiating a 15 deg ramp to to drive onto a 4-ft-high loading
platform. The truck weighs 5 000 lb and has a 42-in wheelbase. Design two (one for each side)
1-ft-wide ramps of steel to have a safety factor of 3 in the worst case of loading as the truck travels
up them. Minimize the weight of the ramps by using a sensible cross-sectional geometry. Choose
an appropriate steel or aluminum alloy.
Given:
Ramp angle
θ  15 deg
Platform height h  4  ft
Truck wheelbase Lt  42 in
Ramp width
Truck weight
w  12 in
W  5000 lbf
Assumptions: 1. The worst case is when the truck CG is located at the center of the beam's span.
2. Use a coordinate frame that has the x-axis along the long axis of the beam.
3. Ignore traction forces and the weight components along the x-axis of the beam.
4. There are two ramps, one for each side of the forklift.
See Figure 5-28 and Mathcad file P0528.
Solution:
L
b
a
CG a
y
CG b
R1

Fa
Wa
Fb
x
Wb
R2
FIGURE 5-28A
Dimensions and Free Body Diagram for Problem 5-28
1. From Problem 3-28 the maximum bending moment in the ramp occurs at the rear wheel of the truck and is
Mmax  8324 ft  lbf
Mmax  99888 in lbf
2. The bending stress is the only stress component present and is, therefore, also the only nonzero principal stress
and is also the von Mises stress. The governing design equations then are
σ' =
Mmax
Z
and
Ns =
Sy
σ'
3. The approach will be to 1) choose a suitable factor of safety, 2) choose a suitable material and determine its yiel
strength, 3) from the equations above determine the required value of the section modulus, 4) choose an
appropriate cross-section for the ramp, and 5) determine the dimensions of the cross-section.
4. The following design choices have been made for this problem:
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-28-2
Design factor of safety
Nsd  3
Material
7075 Aluminum, heat treated
Yield strength
S y  73 ksi
5. Solve the design equations for the minimum section modulus, Z.
Nsd  Mmax
Z 
3
Z  4.105 in
Sy
This is the minimum allowable value of the section modulus.
6. Assume a channel section such as that shown in Figure 5-28B. To keep it simple, let the thickness of the
flanges and web be the same. Choose 1/2-in thick plate, which is readily available. Then, t  0.50 in
A ( h )  w t  2  t ( h  t)
7. The cross-sectional area of the ramp is
cg( h ) 
8. The distance to the CG is
 w t 2
1
A (h)

 2
2
 t h  t
2


9. The moments of inertia of the web and a flange are
Iweb( h ) 
Ifl ( h ) 
w t
3
12
 w t  cg( h ) 

t ( h  t)
3
12
t


Flange
2
Web
2
h  t
 h  t  cg( h ) 

2 

t
2
I ( h )  Iweb( h )  2  Ifl ( h )
h
11. The maximum stress will occur in the flange at the top
and is compressive. The distance from the centroid up to
the top of the flange is
w
c( h )  h  cg( h )
12. Using the known section modulus, solve
for the unknown flange height, h. Guess
h  1  in
FIGURE 5-28B
Channel Section for Problem 5-28
Given
Z=
I (h)
c( h )
h  Find ( h )
h  3.843 in
Round this up to
h  4.00 in
13. Summarizing, the ramp design dimensions are:
Width
w  12.00 in
Flange height
h  4.00 in
Shape
channel
Thickness
t  0.5 in
Material
7075 aluminum
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-29-1
PROBLEM 5-29
Statement:
A differential element is subected to the stresses given below and a ductile material has the
strengths given below. Calculate the safety factor and draw 1-3 diagrams of each theory
showing the stress state using:
(a) Maximum shear-stress theory, and
(b) Distortion-energy theory.
Given:
Principal stresses
σ1  10 ksi
σ2  0  ksi
σ3  20 ksi
Material properties
S ut  50 ksi
S y  40 ksi
S uc  50 ksi
Solution:
See Figure 5-29 and Mathcad file P0529.
1. Calculate the slope of load line. (The load line is the line from the origin through the stress point.)
m 
σ3
m  2
σ1
2. The safety factor equation for the distortion-enrgy theory is the same regardless of which quadrant the load line
falls in. However, the equation for the maximum shear-stress factor of safety is different for each of the three
quadrants that the load line (1st, 3rd, or 4th) can fall in. In this case, the load line falls in the 4th quadrant. The
factors of safety are:
(a) Maximum shear-stress theory
Sy
Na 
Na  1.3
σ1  σ3
3
(b) Distortion energy theory
40
σ1  σ1 σ3  σ3
σ'  26.5 ksi
Nb 
Sy
σ'
(a) Maximum shear
stress boundary
2
Nb  1.5
3. Plot the 1-3 diagram showing the safe-fail
boundaries, the stress state point (10 ksi, -20
ksi) and the load line. Note that if 1 > 3 ,
then only that area on the graph that is to the
right of and below the diagonal line can
contain valid stress points. The factor of
safety is the distance along the load line from
the origin to the intersection of the load line
with the failure boundary, divided by the
distance from the origin to the stress point.
Since the distance from the origin to the
distortion-energy boundary is greater than the
distance to the maximum shear-stress
baoundary, its factor of safety is greater.
30
(b) Distortion energy
boundary
MINIMUM NONZERO PRINCIPAL STRESS, KSI
σ' 
2
20
10
0
1
sy
-10
(10,-20)
-20
-30
Stress states at
which failure
will occur
-sy
-40
Load Line
-50
-60
-40
-30
-20
-10
0
10
20
30
40
50
MAXIMUM PRINCIPAL STRESS, KSI
FIGURE 5-29
1 -  3 Diagram for Problem 5-29
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-30-1
PROBLEM 5-30
Statement:
A differential element is subected to the stresses and strengths given below. Calculate the
safety factor and draw 1-3 diagrams of each theory showing the stress state using:
(a) Coulomb-Mohr theory, and
(b) Modified Mohr theory.
Given:
Principal stresses
σ1  10 ksi
σ2  0  ksi
σ3  20 ksi
Material properties
S ut  50 ksi
S y  40 ksi
S uc  90 ksi
Solution:
See Figure 5-30 and Mathcad file P0530.
1. Calculate the slope of load line. (The load line is the line from the origin through the stress point.)
m 
σ3
σ1
m  2
2. The safety factor equation for both theories is different for each quadrant the load line falls in. The equation for
the modified Mohr factor of safety is different for each of the two regions in the 4th quadrant that the load line can
fall in. In this case, the load line falls in the 4th quadrant, below the -1 slope line.. The factors of safety are:
(a)Coulomb-Mohr theory
50
40
S uc
Nb  3.2
3. Plot the 1-3 diagram showing the safe-fail
boundaries, the stress state point (10 ksi, -20
ksi) and the load line. Note that if 1 > 3 ,
then only that area on the graph that is to the
right of and below the diagonal line can
contain valid stress points. The factor of
safety is the distance along the load line from
the origin to the intersection of the load line
with the failure boundary, divided by the
distance from the origin to the stress point.
Since the distance from the origin to the
modified Mohr boundary is greater than the
distance to the Coulomb-Mohr boundary, its
factor of safety is greater.
30
MINIMUM NONZERO PRINCIPAL STRESS, KSI
 S uc  S ut 
 S
  σ1  σ3
ut


Na  2.4
S uc σ1  S ut σ3
3
(b) Modified Mohr theory
Nb 
S uc S ut
Na 
20
(a) Coulomb-Mohr
boundary
10
1
0
-10
(10,-20)
-20
-30
-40
Stress states at
which failure
will occur
-50
-60
-S
(b) Modified Mohr
boundary
-70
Load Line
-80
-90
-S
ut
uc
-100
-100 -90 -80 -70 -60 -50 -40 -30 -20 -10
0
10
20
30
40
50
MAXIMUM PRINCIPAL STRESS, KSI
FIGURE 5-30
1 -  3 Diagram for Problem 5-30
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-31-1
PROBLEM 5-31
Statement:
Design a jack-stand in a tripod configuration that will support 2 tons of load with a safety factor of
3. Use SAE 1020 steel and minimize its weight.
Solution:
This open-ended design problem has many valid solutions that are left to the student.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-32-1
PROBLEM 5-32
Statement:
A part has the combined stress state and strengths given below. Choose an appropriate failure
theory based on the given data, find the effective stress and factor of safety against static failure.
Given:
Stresses: σx  10 ksi
Strengths:
Solution:
σy  5  ksi
S y  18 ksi
τxy  4.5 ksi
S ut  20 ksi
S uc  80 ksi
See Mathcad file P0532.
1.
Because S uc is greater than S ut, this is an uneven material, which is characteristic of a brittle material. Therefore,
use the modified Mohr theory.
2.
Find the maximum shear stress and principal stresses that result from this combination of applied stresses using
equations 4.6.
Maximum shear stress
τmax 
Principal stresses
σ1 
σ2 
2
 σx  σy 
2

  τxy
2


σx  σy
2
σx  σy
2
τmax  5.148 ksi
 τmax
σ1  12.648 ksi
 τmax
σ2  2.352 ksi
σ3  0  psi
3.
4.
Find the Dowling factors C1, C2, C3 using equations 5.12b:
C1 
1
C2 
1
C3 
1
2
2
2

S uc  2  S ut

S uc

S uc  2  S ut

S uc

S uc  2  S ut

S uc
  σ1  σ2 
  σ2  σ3 
  σ3  σ1 

  σ1  σ2
C1  8.898 ksi


  σ2  σ3
C2  1.764 ksi


  σ3  σ1
C3  9.486 ksi

Then find the largest of the six stresses C1, C2, C3 , 1, 2, 3:
C
  1  
  C2  
 C 
3
σeff  max   
  σ1  
 
  σ2  
  σ3  
σeff  12.6 ksi
which is the modified-Mohr effective stress.
5.
The safety factor can now be found using equation 5.12d.
N 
S ut
σeff
N  1.6
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-33a-1
PROBLEM 5-33a
Statement:
For the bracket shown in Figure P5-14 and the data in row a of Table P5-3, determine the von
Mises stresses at points A and B.
Solution:
See Mathcad file P0533a.
1.
From Problem 4-33a the principal stresses at point A are
σ1  21.46  MPa
2.
2
σ1  σ1 σ3  σ3
2
σ'A  30.2 MPa
From Problem 4-33a the principal stresses at point B are
σ1  16.13  MPa
4.
σ3  13.08  MPa
Use equation (5.7c) to find the von Mises stress at point A.
σ'A 
3.
σ2  0  MPa
σ2  0  MPa
σ3  16.13  MPa
Use equation (5.7c) to find the von Mises stress at point B.
σ'B 
2
σ1  σ1 σ3  σ3
2
σ'B  27.9 MPa
F
y
A
B
T
T
x
M
L
R
FIGURE 5-33
Free Body Diagram of Tube for Problem 5-33
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-34a-1
PROBLEM 5-34a
Statement:
Calculate the safety factor for the bracket in Problem 5-33 using the distortion energy, the
maximum shear stress, and the maximum normal-stress theories. Comment on their
appropriateness. Assume a ductile material strength as given below.
Given:
Yield strength
Solution:
See Mathcad file P0534a.
1.
S y  400  MPa
From Problem 4-33a the principal stresses at point A are
σ1A  21.46  MPa
2.
σ2A  0  MPa
σ3A  13.08  MPa
Using the two nonzero stresses, the slope of the load line on a 1-3 graph is
m 
σ3A
m  0.61
σ1A
This intersects the failure boundaries in the fourth quadrant.
3.
Calculate the von Mises effective stress at point A using equation (5.7c).
σ'A 
4.
5.
2
2
σ1A  σ1A σ3A  σ3A
Determine the factor of safety at point A
Distortion energy
NADE 
Maximum shear stress
NAMS 
Maximum normal stress
NANS 
Sy
NADE  13.2
σ'A
Sy
σ1A  σ3A
Sy
σ1A
NAMS  11.6
NANS  18.6
From Problem 4-33a, the principal stresses at Point B are
σ1B  16.13  MPa
6.
σ'A  30.205 MPa
σ2B  0  MPa
σ3B  16.13  MPa
Using the two nonzero stresses, the slope of the load line on a 1-3 graph is
m 
σ3B
m  1
σ1B
This intersects the failure boundaries in the fourth quadrant.
7.
Calculate the von Mises effective stress at point A using equation (5.7c).
σ'B 
8.
2
2
σ1B  σ1B σ3B  σ3B
σ'B  27.938 MPa
Determine the factor of safety at point B
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
9.
Distortion energy
NBDE 
Maximum shear stress
NBMS 
Maximum normal stress
NBNS 
5-34a-2
Sy
NBDE  14.3
σ'B
Sy
σ1B  σ3B
Sy
σ1B
NBMS  12.4
NBNS  24.8
Whichever theory is used, the critical point (lowest factor of safety) is point A. The distortion energy theory
should be used because experimental data follows its failure boundary more nearly than the maximum shear
stress in all quadrants. Using the maximum normal stress theory would give an overestimate of the actual safety
factor.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-35a-1
PROBLEM 5-35a
Statement:
Calculate the safety factor for the bracket in Problem 5-33 using the Coulomb-Mohr and the
modified Mohr effective stress theories. Comment on their appropriateness. Assume a brittle
material strength as given below.
Given:
Tensile strength
Solution:
See Mathcad file P0535a.
1.
S ut  350  MPa
S uc  1000 MPa
From Problem 4-33a the principal stresses at point A are
σ1A  21.46  MPa
2.
Compressive strength
σ2A  0  MPa
σ3A  13.08  MPa
Using the two nonzero stresses, the slope of the load line on a 1-3 graph is
m 
σ3A
m  0.61
σ1A
This intersects the failure boundaries in the fourth quadrant. For the Coulomb-Mohr diagram (see Figure 5-9
in the text) there is a single, straight line in this quadrant. For the modified-Mohr theory, the load line will
intersect the boundary at a point similar to B' in Figure 5-11 in the text.
3.
4.
Determine the factor of safety at point A
Coulomb-Mohr
NACM 
Modified-Mohr
NAMM 
S uc σ1A  S ut σ3A
S ut
σ1A
NACM  13.4
NAMM  16.3
From Problem 4-33a, the principal stresses at Point B are
σ1B  16.13  MPa
5.
S ut S uc
σ2B  0  MPa
σ3B  16.13  MPa
Using the two nonzero stresses, the slope of the load line on a 1-3 graph is
m 
σ3B
m  1
σ1B
This intersects the failure boundaries in the fourth quadrant. For the Coulomb-Mohr diagram (see Figure 5-9
in the text) there is a single, straight line in this quadrant. For the modified-Mohr theory, the load line will
intersect the boundary at the point (S ut -S ut) Figure 5-11 in the text.
6.
7.
Determine the factor of safety at point B
Coulomb-Mohr
NBCM 
Modified-Mohr
NBMM 
S ut S uc
S uc σ1B  S ut σ3B
S ut
σ1B
NBCM  16.1
NBMM  21.7
Whichever theory is used, the critical point (lowest factor of safety) is point A. The modified-Mohr theory
should be used because experimental data follows its failure boundary more nearly than the Coulomb-Mohr
when the slope of the load line is in the fourth quadrant. Using the Coulomb-Mohr would give an underestimat
of the actual safety factor.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
8.
5-35a-2
Calculating factor of safety using Modified Mohr and equations (5.12c, d, and e)
Point A
C1 
1
C2 
1
C3 
1
2
2
2

S uc  2  S ut

S uc

S uc  2  S ut

S uc

S uc  2  S ut

S uc
  σ1A  σ2A 
  σ2A  σ3A 
  σ3A  σ1A 
1 is maximum so

  σ1A  σ2A


  σ2A  σ3A


  σ3A  σ1A

N 
S ut
C1  13.9 MPa
C2  4.6 MPa
C3  18.5 MPa
N  16.3
σ1A
Point B
C1 
1
C2 
1
C3 
1
2
2
2

S uc  2  S ut

S uc

S uc  2  S ut

S uc

S uc  2  S ut

S uc
  σ1B  σ2B 
  σ2B  σ3B 
  σ3B  σ1B 
1 is maximum so

  σ1B  σ2B


  σ2B  σ3B


  σ3B  σ1B

N 
S ut
σ1B
C1  10.5 MPa
C2  5.6 MPa
C3  16.1 MPa
N  21.7
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-36a-1
PROBLEM 5-36a
Statement:
For the bracket shown in Figure P5-14 and the data in row a of Table P5-3, redo Problem 5-33
considering the stress concentration at points A and B. Assume a stress concentration factor of
2.5 in both bending and torsion.
Given:
Factors of safety:
Bending
Solution:
1.
2.
4.
Kfs  2.5
Torsion
See Mathcad file P0536a.
From Problem 4-36a the principal stresses at point A are
σ1  53.6 MPa
σ2  0  MPa
σ3  32.7 MPa
Use equation (5.7c) to find the von Mises stress at point A.
σ'A 
3.
Kf  2.5
2
σ1  σ1 σ3  σ3
2
σ'A  75.5 MPa
From Problem 4-36a the principal stresses at point B are
σ1  41.3 MPa
σ2  0  MPa
σ3  41.3 MPa
Use equation (5.7c) to find the von Mises stress at point B.
σ'B 
2
σ1  σ1 σ3  σ3
2
σ'B  71.5 MPa
F
y
A
B
T
T
x
M
L
R
FIGURE 5-36
Free Body Diagram of Tube for Problem 5-36
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-37-1
PROBLEM 5-37
Statement:
Given:
S y  700  MPa
A semicircular, curved beam as shown in Figure 5-37 has the dimensions given below. For a load
pair F = 14 kN applied along the diameter, find the safety factor at the inner and outer fibers:
(a) If the beam is a ductile material with Sy = 700 MPa,
(b) If the beam is a cast-brittle material with Sut = 420 MPa, Suc = 1200 MPa.
(b) Tensile strength
Solution:
1.
w
(a) Yield strength
S ut  420  MPa
Compressive strength
S uc  1200 MPa
See Figure 5-37 and Mathcad file P0537.
From Problem 4-37, the stresses at the inside radius and
outside radius are:
Inside
σi  409.9  MPa
Outside
σo  273.2  MPa
F
od
id
F
(a) Entire Beam
These are the only stress components present on their
respective surfaces so they are also principal stresses. Thus,
σ1i  409.9  MPa
σ2i  0  MPa
σ3i  0  MPa
σ1o  0  MPa
σ2o  0  MPa
σ3o  273.2  MPa
F
M
F
Part (a)
2.
rc
Use the distortion energy theory for the ductile material.
3. Since 1 is the only nonzero principal stress, it is also the
von Mises effective stress,
(b) Critical Section
FIGURE 5-37
Free Body Diagrams for Problem 5-37
4.
σ'i  σ1i
σ'i  409.9 MPa
σ'o  σ3o
σ'o  273.2 MPa
The factor of safety against a static failure for this ductile material is
Inside surface
Nai 
Outside surface
Nao 
Sy
σ'i
Sy
σ'o
Nai  1.7
Nao  2.6
Part (b)
5.
Use the modified-Mohr theory for the brittle material.
6.
The load line on the 1-3 graph for the inside surface is along the positive 1 axis. In this case, the factor of
safety equation simplifies to
Inside surface
7.
Nbi 
S ut
σ1i
Nbi  1.0
The load line on the 1-3 graph for the outside surface is along the negative 3 axis. In this case, the factor of
safety equation simplifies to
Outside surface
Nbo 
S uc
σ3o
Nbo  4.4
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-38-1
PROBLEM 5-38
Statement:
Assume that the curved beam of Problem 5-37 has a crack on its inside surface of half-width a = 2
mm and a fracture toughness of 50 MPa-m0.5. What is its safety factor against sudden fracture?
Given:
Outside diameter
Width of section
od  150  mm
t  25 mm
Inside diameter
Load
id  100  mm
F  14 kN
Half crack length
a  2  mm
Fracture toughness
Kc  50 MPa m
Solution:
See Figure 5-38 and Mathcad file P0538.
1.
From Problem 4-37, the nominal stress at the inside radius is:
Nominal inside stress
σi  409.9  MPa
2.
Calculate the half-width of the beam.
3.
Calculate the geometry and stress intensity factors.
π a 
β  sec 
β  1.016

 2 b 
K  β  σi π a
4.
b  0.5 t
b  12.5 mm
K  33.01 MPa m
Determine the factor of safety against sudden fracture failure
NFM 
Kc
K
NFM  1.5
w
F
od
id
F
(a) Entire Beam
F
M
F
rc
(b) Critical Section
FIGURE 5-38
Free Body Diagrams for Problem 5-38
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-39-1
PROBLEM 5-39
Statement:
Consider the failed 260-in dia by 0.73-in wall rocket case of Figure 5-14. The steel had S y = 240 k
and a fracture toughness Kc = 79.6 ksi-in 0.5. It was designed for an internal pressure of 960 psi but
failed at 542 psi. Failure was attributed to a small crack that precipitated a sudden, brittle,
fracture-mechanics failure. Find the nominal stress in the wall and the yielding safety factor at the
failure conditions and estimate the size of the crack that caused it to explode. Assume b = 1.0.
Given:
Solution:
1.
2.
3.
Case diameter
d  260  in
Fracture toughness
Kc  79.6 ksi in
Wall thickness
t  0.73 in
Design pressure
p d  960  psi
Yield strength
S y  240  ksi
Failure pressure
p f  542  psi
See Mathcad file P0539.
Find the nominal stress in the wall. The ratio of the wall thickness to the radius of the case is such that we can
use thin-wall theory. Thus
Case radius
r  0.5 d
Tangential stress
σt 
r  130 in
pd  r
σt  171.0 ksi
t
pd  r
Axial stress
σa 
Radial stress
σr  0  psi
σa  85.5 ksi
2 t
Find the yielding safety factor at the failure conditions. Since, for these directions, there are no shear stresses
present, these are the principal stresses. The von Mises stress is
Von Mises stress
σ' 
Factor of safety
against yielding
Ns 
2
σt  σt σa  σa
Sy
σ'
2
σ'  148.1 ksi
Ns  1.6
Estimate the size of the crack that caused it to explode.
Tangential stress
Axial stress
σt 
σa 
pf  r
t
pf  r
2 t
σt  96.5 ksi
σa  48.3 ksi
(a) Assume that the crack was longitudinal (growing in the axial direction)
Nominal stress
σnom  σt
Stress intensity
factor
K = σnom a  π
σnom  96.5 ksi
Setting the stress intensity factor equal to the fracture toughness of the material and solving for the crack length
 Kc2 


 σ 2 π 
 nom 
Half-length
a 
Crack length
2  a  0.433 in
a  0.216 in
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-39-2
(b) Assume that the crack was tangential (growing in the tangential direction)
Nominal stress
σnom  σa
Stress intensity
factor
K = σnom a  π
σnom  48.3 ksi
Setting the stress intensity factor equal to the fracture toughness of the material and solving for the crack length
Half-length
 Kc2 

a  
 σ 2 π 
 nom 
Crack length
2  a  1.732 in
a  0.866 in
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-40-1
PROBLEM 5-40
Statement:
Redesign the roll support of Problem 5-8 to be like that shown in Figure P5-16. The stub mandrels
insert to 10% of the roll length at each end. Design dimension a for a factor of safety of 2. See
Problem 5-8 for additional data.
(a) The beam is a ductile material with S y = 300 MPa
(b) The beam is a cast-brittle material with S ut = 150 MPa, S uc = 570 MPa.
Given:
Paper roll dimensions:
OD  1.50 m
ID  0.22 m
Material properties:
Yield strength
Lroll  3.23 m
Tensile strength S ut  150  MPa
Comp strength
Roll density
S y  300  MPa
3
ρ  984  kg m
S uc  570  MPa
Ns  2
Factor of safety
Assumptions: 1. The paper roll's weight creates a concentrated load acting at the tip of the mandrel.
2. The base of the mandrel (the portion that inserts into the stanchion) is solid and fits tightly into
the stanchion. Therefore, the mandrel can be treated as a cantilever beam.
3. The length of ther mandrel base is b  100  mm.
Solution:
1.
2.
For the assumptions made, it is not necessary to
determine the stress distribution on the mandrel base
inside the stanchion. From Figure 5-40, we see that we
can determine the diameter a by applying the beam
stress equation at the section where the mandrel
transitions from the base to the full diameter.
π
4

2
2

 OD  ID  Lroll  ρ  g
y
x
a
M1
Lm
b
W  53.9 kN
R
F  0.5 W
F  26.95 kN
FIGURE 5-40
Lm  0.1 Lroll
Lm  323 mm
Free Body Diagram used in Problem 5-40
From Figure 5-40, the maximum internal bending moment occurs at x = 0 and is
Mmax  F  Lm
4.
F
Determine the weight of the roll, the load on each
support, and the length of the mandrel.
W 
3.
See Figure 5-40 and Mathcad file P0540.
Mmax  8.704 kN  m
The bending stress will be a maximum at the top or bottom of the mandrel at a section through x = 0.
σmax =
Mmax a
2 I
where
I=
π a
4
64
There are no other stress components at this point so σmax = σ1 and
σ2  0  MPa
5.
σ3  0  MPa
For the ductile material of part (a), the maximum principal stress is also the von Mises stress so
σmax = σ' =
32 Mmax
π a
3
=
Sy
Ns
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-40-2
1
5.
Solving for a,
 32 Ns F  Lm 
a  

 π S y 
Round this to
a  84 mm
3
a  83.922 mm
for the ductile material of part (a)
For the brittle material of part (b), the load line on the 1-3 diagram is along the positive 1 axis where both
brittle material failure theories have the same boundary, which is 1 = S ut. Thus, for the brittle case of part (b),
σmax = σ1 =
32 Mmax
π a
3
=
S ut
Ns
1
Solving for a,
 32 Ns F  Lm 
a  

 π S ut 
Round this to
a  106  mm
3
a  105.735 mm
for the brittle material of part (b)
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-41-1
PROBLEM 5-41
Statement:
A 10-mm ID steel tube carries liquid at 7 MPa. The steel has S y = 400 MPa Determine the safety
factor for the wall if its thickness is: a) 1 mm, b) 5 mm.
Given:
Yield strength
Assumption:
The tubing is long therefore the axial stress is zero.
Solution:
See Mathcad file P0541.
t  1  mm
(a) Wall thickness is
1.
S y  400  MPa
From Problem 4-41, this is a thick wall cylinder and the principal stresses are:
σ1a  38.82  MPa
2.
2
2
σ1a  σ1a σ3a  σ3a
Sy
Na  9.4
σ'a
t  5  mm
(b) Wall thickness is
From Problem 4-41, this is a thick wall cylinder and the principal stresses are:
σ1b  11.67  MPa
5.
σ2b  0  MPa
σ3b  7.00 MPa
Calculate the von Mises effective stress using equation (5.7c).
σ'b 
6.
σ'a  42.752 MPa
Using the distortion energy theory, the factor of safety is
Na 
4.
σ3a  7.00 MPa
Calculate the von Mises effective stress using equation (5.7c).
σ'a 
3.
σ2a  0  MPa
2
2
σ1b  σ1b σ3a  σ3b
σ'b  16.336 MPa
Using the distortion energy theory, the factor of safety is
Nb 
Sy
σ'b
Nb  24.5
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-42-1
PROBLEM 5-42
Statement:
A cylindrical tank with hemispherical ends is required to hold 150 psi of pressurized air at room
temperature. The steel has S y = 400 MPa. Determine the safety factor if the tank diameter is 0.5 m
with 1 mm wall thickness, and its length is 1 m.
Given:
Yield strength
Solution:
See Mathcad file P0542.
1.
S y  400  MPa
From Problem 4-42, the maximum principal stresses in the wall are
σ1  259  MPa
σ2  129  MPa
σ' 
2
σ1  σ1 σ2  σ2
σ3  0  MPa
2
σ'  224.301 MPa
2.
The von Mises stress is
3.
Using the distortion-energy theory, the factor of safety against a static failure is
Ns 
Sy
σ'
Ns  1.8
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-43-1
PROBLEM 5-43
Statement:
The paper rolls in Figure P5-17 are 0.9-m OD by 0.22-m ID by 3.23-m long and have a density of
984 kg/m3. The rolls are transfered from the machine conveyor (not shown) to the forklift truck
by the V-linkage of the off-load station, which is rotated through 90 deg by an air cylinder. The
paper then rolls onto the waiting forks of the truck. The forks are 38-mm thick by 100-mm wide
by 1.2-m long and are tipped at a 3-deg angle from the horizontal and have Sy = 600MPa. Find
the safety factor for the two forks on the truck when the paper rolls onto it under two different
conditions (state all assumptions):
(a) The two forks are unsupported at their free end.
(b) The two forks are contacting the table at point A.
F
Given:
S y  600  MPa
Yield strength
Assumptions: 1. The greatest bending moment will occur
when the paper roll is at the tip of the fork
for case (a) and when it is midway between
supports for case (b).
2. Each fork carries 1/2 the weight of a
paper roll.
3. For case (a), each fork acts as a cantilever
beam (see Appendix B-1(a)).
4. For case (b), each fork acts as a beam
that is built-in at one end and
simply-supported at the other.
Solution:
L fork
t
R1
Case (a), Cantilever Beam
0.5 L fork
F
t
L fork
See Figure 5-43 and Mathcad file P0543.
R1
1.
From Problem 4-43, the maximum stresses in the
forks are:
Case (a)
M1
R2
M2
Case (b), Fixed-Simply Supported Beam
FIGURE 5-43
σa  464.8  MPa
Free Body Diagrams used in Problem 5-43
at the base of the fork.
Case (b)
σb  87.2 MPa
also at the base of the fork.
Since there are no other stress components present, these are also the maximum principal stresses and the
von Mises stresses. Thus, σ'a  σa and σ'b  σb.
Case (a)
2.
The factor of safety against a static failure is
Nsa 
Sy
σ'a
Nsa  1.3
Case (b)
3.
The factor of safety against a static failure is
Nsb 
Sy
σ'b
Nsb  6.9
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-44-1
PROBLEM 5-44
Statement:
Determine a suitable thickness for the V-links of the off-loading station of Figure P5-17 to limit
their deflections at the tips to 10-mm in any position during their rotation. Two V-links support
the roll, at the 1/4 and 3/4 points along the roll's length, and each of the V arms is 10-cm wide by
1-m long. What is their safety factor against yielding when designed to limit deflection as
above?
Given:
Roll OD
OD  0.90 m
Arm width
wa  100  mm
Roll ID
ID  0.22 m
Arm length
La  1000 mm
Roll length
Lroll  3.23 m
Max tip deflection
δtip  10 mm
Roll density
Yield strength
ρ  984  kg m
Mod of elasticity
E  207  GPa
3
S y  400  MPa
Assumptions: 1. The maximum deflection on an arm will occur just after it begins the transfer and just before it
completes it, i.e., when the angle is either zero or 90 deg., but after the tip is no longer supported b
the base unit.
2. At that time the roll is in contact with both arms ("seated" in the V) and will remain in that state
throughout the motion. When the roll is in any other position on an arm the tip will be supported.
3. The arm can be treated as a cantilever beam with nonend load.
4. A single arm will never carry more than half the weight of a roll.
5. The pipe to which the arms are attached has OD = 160 mm.
Solution:
See Figure 5-44 and Mathcad file P0544.
1. Determine the weight of the roll and the load on each
V-arm.
W 

4
π
2
2

 OD  ID  Lroll  ρ  g
450
W  18.64  kN
F  0.5 W
F  9.32 kN
2. From Appendix B, Figure B-1, the tip deflection of a
cantilever beam with a concentrated load located at a
distance a from the support is
ymax =
F a
2
6  E I
 ( a  3  L)
1000 = L
370 = a
where L is the beam length and I is the
cross-section moment of inertia. In this case
F
3
I=
3. Setting
w a t a
M
12
ymax = δtip
F
a  370  mm
and
FIGURE 5-44
Free Body Diagram used in Problem 5-44
substituting for I and solving for ta
1
 2 F  a2  3 La  a 
ta 


E δtip  wa


Let the arm thickness be
3
ta  31.889 mm
ta  32 mm
4. The maximum bending stress in the arm will be at its base where it joins the 160-mm-dia pipe. The bending
moment, moment of inertia, and distance to the outside fiber at that point are:
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
Bending moment
Moment of inertia
Distance to outer fiber
M  a  F
I 
wa ta
5-44-2
M  3449 N  m
3
12
c  0.5 t a
5
I  2.731  10  mm
4
c  16 mm
5. The bending stress, which is also the von Mises stress, is
σ' 
M c
I
σ'  202.1  MPa
6. Using the distortion-energy theory, the factor of safety against a static failure is
Ns 
Sy
σ'
Ns  2.0
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-45-1
PROBLEM 5-45
Statement:
Determine the safety factor based on critical load on the air cylinder rod in Figure P5-17 if the
crank arm that rotates it is 0.3 m long and the rod has a maximum extension of 0.5 m. The
25-mm-dia rod is solid steel with a yield strength of 400 MPa. State all assumptions.
Given:
Rod length
Rod diameter
L  500  mm
d  25 mm
Young's modulus
Yield strength
E  207  GPa
S y  400  MPa
Assumptions: 1. The rod is a fixed-pinned column.
2. Use a conservative value of 1 for the end factor (see Table 4-7 in text).
Solution:
See Problems 4-45, 4-47, and Mathcad file P0545.
1.
From Problem 4-45, the critical load on the air cylinder rod is
Pcr  134.8  kN
2.
From Problem 4-47, the maximum load on the air cylinder rod is
F  46.47  kN
3.
The factor of safety against a buckling failure is Nbuck 
Pcr
F
Nbuck  2.9
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-46-1
PROBLEM 5-46
Statement:
The V-links of Figure P5-17 are rotated by the crank arm through a shaft that is 60 mm dia by
3.23 m long. Determine the maximum torque applied to this shaft during motion of the V-linkage
and find the static safety factor against yielding for the shaft if its S y = 400 MPa. See Problem
5-43 for more information.
y
Given:
Yield strength
S y  400  MPa
Assumptions: The greatest torque will occur
when the link is horizontal and the
paper roll is located as shown in
Figure P5-17 or Figure 5-46.
Solution:
P0546.
See Figure 5-46 and Mathcad file
1. From Problem 4-46, the maximum torsional stress
in the shaft is
W
τmax  197.88 MPa
T
2. Using the distortion-energy theory, the factor of
safety against static yielding is
Ry
60-mm-dia shaft
Ns 
Sy
3  τmax
Ns  1.2
450.0
FIGURE 5-46
Free Body Diagram used in Problem 5-46
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-47-1
PROBLEM 5-47
Statement:
Determine the maximum forces on the pins at each end of the air cylinder of Figure P4-17.
Determine the safety factor for these pins if they are 30-mm dia and in single shear. S y = 400 MPa.
Given:
Paper roll dimensions
OD  0.90 m
ID  0.22 m
Pin diameter
Yield strength
d  30 mm
S y  400  MPa
Lroll  3.23 m
3
ρ  984  kg m
Roll density
Assumptions: 1. The maximum force in the cylinder rod occurs when the transfer starts.
2. The cylinder and rod make an angle of 8 deg to the horizontal at the start of transfer.
3. The crank arm is 300 mm long and is 45 deg from vertical at the start of transfer.
4. The cylinder rod is fully retracted at the start of the transfer. At the end of the transfer it will
have extended 500 mm from its initial position.
Solution:
See Figure 4-47 and Mathcad file P0447.
1. Determine the weight of the roll on the
forks.
W 

4
π
2
2
y

 OD  ID  Lroll  ρ  g
W  18.64 kN
2. From the assumptions and
Figure 4-47, the x and y
distances from the origin to
point A are,
Rax  300  cos( 45 deg)  mm
W
Ray  300  sin( 45 deg)  mm
Rx
Rax  212.132 mm
x
212.1
Ry
A
Ray  212.132 mm
F
8°
212.1
450.0
3. From Figure 4-47, the x
distance from the origin to
point where W is applied is,
FIGURE 4-47
Free Body Diagram at Start of Transfer for V-link of Problem 4-47
Rwx 
OD
2
Rwx  450 mm
4. Sum moments about the pivot point and solve for the compressive force in the cylinder rod.
W  Rwx  F  Rax sin( 8  deg)  F  Ray cos( 8  deg) = 0
F 
W  Rwx
Ray cos( 8  deg)  Rax sin( 8  deg)
F  46.469 kN
This is the shear force in the pins
5. Determine the cross-sectional area of the pins and the direct shear stress.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
Shear area
Shear stress
A 
τ 
π d
2
4
F
A  706.858 mm
5-47-2
2
τ  65.7 MPa
A
6. Using the distortion-energy theory, the factor of safety against a static yielding failure is
Ns 
Sy
3 τ
Ns  3.5
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-48-1
PROBLEM 5-48
Statement:
Figure P5-18 shows an exerciser for a 100-kg wheelchair racer. The wheel chair has 65 cm dia
drive wheels separated by a 70-cm track width. Two free-turning rollers on bearings support the
rear wheels. The lateral movement of the chair is limited by the flanges. Design the 1-m-long
rollers as hollow tubes of aluminum (select alloy) to minimize the height of the platform and also
limit the roller deflections to 1 mm in the worst case. Specify suitable sized steel axles to support
the tubes on bearings. Calculate all significant stresses.
Given:
Mass of chair
M  100  kg
Wheel diameter d w  650  mm
Track width
T  700  mm
Aluminum
Ea  71.7 GPa
Roller length
Lr  1000 mm
Steel
Es  207  GPa
Assumptions: 1. The CG of the chair with rider is
sufficiently close to the rear wheel that all of
the weight is taken by the two rear wheels.
2. The small camber angle of the rear wheels
does not significantly affect the magnitude
of the forces on the rollers.
3. Both the aluminum roller and the steel axle
are simply supported. The steel axles that
support the aluminum tube are fixed in the
mounting block and do not rotate. The
aluminum tube is attached to them by two
bearings (one on each end of the tubes, one
for each axle). The bearings' inner race is
fixed, and the outer race rotates with the
aluminum tube. Each steel axle is considered
to be loaded as a simply supported beam.
Their diameter must be less than the inner
diameter of the tubes to fit the roller
bearings between them.
Solution:
δ  1  mm
Maximum deflection
Modulus elasticity:
W/2
F
F


FIGURE 5-48A
Free Body Diagram of One Wheel
used in Problem 5-48
See Figures 5-48 and Mathcad file P0548.
1. Calculate the weight of the chair with rider.
Weight of chair
W  M  g
W  980.7 N
2. Calculate the forces exerted by the wheels on the rollers (see Figure 5-48A). From the FBD of a wheel, summing
vertical forces
2  F  cos( θ ) 
Let
θ  20 deg
W
2
=0
then
F 
W
4  cos( θ )
F  260.9 N
3. The worst condition (highest moment at site of a stress concentration) will occur when the chair is all the way to
the left or right. Looking at the plane through the roller that includes the forces exerted by the wheels (the FBD is
shown in Figure 5-48B) the reactions R1 and R2 come from the bearings, which are inside the hollow roller and are,
themselves, supported by the steel axle.
4. Solving for the reactions. Let the distance from R1 to F be a  15 mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
 M1
R2 Lr  F  ( a  T )  F  a = 0
 Fy
R1  2  F  R2 = 0
R2 
F  (2 a  T )
5-48-2
700
F
R2  190.5 N
Lr
15
R2
R1
R1  2  F  R2
F
1000
R1  331.3 N
FIGURE 5-48B
5. The maximum bending moment will be
at the right-hand load and will be
Free Body Diagram of One Tube used in Problem 5-48
Mrmax  R2 Lr  ( a  T )
Mrmax  54.3 N  m
Note, if the chair were centered on the roller the maximum moment would be
Mc  F 
Lr  T
Mc  39.1 N  m
2
and this would be constant along the axle between the two loads, F.
6. Note that the bearing positions are fixed regardless of the position of the chair on the roller.
Because of symmetry,
Ra1  R1
Ra1  331.3 N
Ra2  R2
Ra2  190.5 N
1000
65
R1
7. The maximum bending moment
occurs at R1 and is for b  65 mm
R2
R a1
R a2
1130
Mamax  Ra1 b
FIGURE 5-48C
Free Body Diagram of One Axle used in Problem 5-48
Mamax  21.5 N  m
8. Determine a suitable axle diameter. Let the factor of safety against yielding in the axle be Nsa  3
9. Tentatively choose a low-carbon steel for the axle, say AISI 1020, cold rolled with S y  393  MPa
10. At the top of the axle under the load R1 there is only a bending stress, which is also the von Mises stress. Set th
stress equal to the yield strength divided by the factor of safety.
σ' =
32 Mamax
π d a
3
=
Sy
Nsa
1
Solving for the axle diameter, d a
 32 Nsa Mamax 
d a  

π S y


Let the axle diameter be
d a  15 mm
3
d a  11.875 mm
made from cold-rolled AISI 1020 steel.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-48-3
11. Suppose that bearing 6302 from Chapter 10, Figure 10-23, page 684 is used. It has a bore of 15 mm and an
OD of 42 mm. Thus, the inside diameter of the roller away from the bearings where the moment is a maximum
will be d i  40 mm. This will provide a 1-mm shoulder for axial location of the bearings.
12. The maximum deflection of the roller will
occur when the chair is in the center of the
roller. For this case the reactions are both equal
to the loads, F (see Figure 4-48D). The
maximum deflection is at the center of the roller.
150
700
F
F
F
15
F
1000
13. Write the load function and then
integrate four times to get the deflection
function.
FIGURE 5-48D
Free Body Diagram of Roller with Chair in the Center.
q(x) = F<x>-1 - F<x - a>-1 - F<x - b>-1 + F<x - L>-1
y(x) = F[<x>3 - <x - a>3 - <x - b>3 + <x - L>3 + C3x]/(6EI)
where
C3 =
1
L
 ( L  a )  a  L 
3
3
3
14. Write the deflection function at x = L/2 for a  150  mm
ymax =
3
 L 3  L
  1  ( L  a) 3  a3  L3


a


 2

6  Ea I  2 
2



 
F
15. Set this equation equal to the allowed deflection  and solve for the required moment of inertia, I.
3
 Lr  3  Lr


1
3
3
3

I 
   
 a    Lr  a   a  Lr 
6  Ea δ  2 
2
2


F
4
I  6.618  10 mm
4
16. Knowing the inside diameter of the tube, solve for the outside diameter.
1
I=
π  4
4
  d o  d i 
64
Round this up to
d o 
 64 I  d 4
 π
i 


4
d o  44.463 mm
d o  46 mm
DESIGN SUMMARY
Axles
Rollers
Material
AISI 1020 steel, cold-rolled
Material
2024-T4 aluminum
Diameter
d a  15 mm
Outside diameter
d o  46 mm
Length
1220 mm
Inside diameter
d i  40 mm
Length
1040 mm
Spacing
c   d w  d o  sin( θ )
c  238 mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
PROBLEM 5-49
Statement:
5-49-1
_____
A part made of ductile steel with Sy = 40 ksi is subjected to a three-dimensional stress state of 1
= -80 ksi, 2 = -80 ksi, 3 = -80 ksi. What is the maximum shear stress? Will the part fail?
Solution:
1.
See Mathcad file P0549.
This is a case of hydrostatic stress. As explained in Section 5.1, the maximum shear stress is zero. Parts loaded
hydrostatically can withstand stresses well in excess of their yield strength. One example of this is that parts on
the ocean floor such as those retrieved from the Titanic are intact and undistorted even though they are
surrounded by water at great pressure.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-50-1
PROBLEM 5-50
_____
Statement:
A component in the shape of a large sheet is to be fabricated from 7075-T651 aluminum, which
has a fracture toughness Kc = 24.2 MPa-m0.5 and a tensile yield strength of 495 MPa. Determine
the largest edge crack that could be tolerated in the sheet if the nominal stress does not exceed
one half the yield strength.
Given:
Fracture toughness
Kc  24.2 MPa m
Yield strength
S y  495  MPa
Solution:
1.
Mathcad file P0550.
Calculate the nominal stress based on the yield strength and the stress level given in the problem statement.
σnom 
2.
0.5
Sy
σnom  247.5 MPa
2
Determine the value of the geometry factor  from the discussion in Section 5.3 for a plate with an edge crack.
β  1.12
3.
Using equation 5.14b, calculate the critical crack length for this material under the given stress condition.
 Kc 
a   

π  β  σnom 
1
2
a  2.4 mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-51-1
PROBLEM 5-51
_____
Statement:
A component in the shape of a large sheet is to be fabricated from 4340 steel, which has a fracture
toughness Kc = 98.9 MPa-m0.5 and a tensile yield strength of 860 MPa. The sheets are inspected
for crack flaws after fabrication, but the inspection device cannot detect flaws smaller than 5 mm.
The part is too heavy as designed. An engineer has suggested that the thickness be reduced and
the material be heat-treated to increase its tensile strength to 1515 MPa, which would result in
decreasing the fracture toughness to 60.4 MPa-m0.5. Assuming that the stress level does not
exceed one half the yield strength, is the suggestion feasible? If not, why not.
Given:
Fracture toughness
Kc1  98.9 MPa m
Kc2  60.4 MPa m
Yield strength
S y1  860  MPa
S y2  1515 MPa
Solution:
1.
0.5
See Mathcad file P0551.
Calculate the nominal stress for the two material conditions based on the yield strength and the stress level
given in the problem statement.
σnom1 
σnom2 
2.
0.5
S y1
σnom1  430 MPa
2
S y2
σnom2  757.5 MPa
2
Determine the value of the geometry factor  from the discussion in Section 5.3 for a large plate.
β  1
3.
Using equation 5.14b, calculate the critical crack length for each material condition under the given stress
condition.
a 1 
2
 Kc1 

π  β  σnom1 
1

 Kc2 
a 2   

π  β  σnom2 
1
4.
a 1  16.8 mm
2  a 1  33.7 mm
a 2  2.0 mm
2  a 2  4.0 mm
2
The suggestion to increase the strength of the material so that its thickness can be decreased to save weight is
not feasible because the critical crack size of the material in the second condition is less than that which can be
detected by the inspection equipment.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-52-1
PROBLEM 5-52
_____
Statement:
A large plate is subjected to a nominal tensile stress of 350 MPa. The plate has a central crack
that is 15.9 mm long. Calculate the stress intensity factor at the tip of the crack.
Given:
Nominal stress
σnom  350  MPa
Crack length
lcrack  15.9 mm
Solution:
1.
See Mathcad file P0552.
Calculate the half-width of the crack
a  0.5 l crack
2.
a  7.95 mm
Determine the value of the geometry factor  from the discussion in Section 5.3 for a plate with an edge crack.
β  1
3.
Using equation 5.14b, calculate the stress intensity factor.
K  β  σnom π a
0.5
K  55.3 MPa m
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-53-1
PROBLEM 5-53
_____
Statement:
A movie scene calls for a stuntman to hang from a rope that is suspended 3 m above a pit of
poisonous spiders. The rope is attached to a glass sheet that is 3000 mm long by 100 mm wide
and 1.27 mm thick. The stuntman knows that the glass sheet contains a central crack with total
length of 16.2 mm that is oriented parallel to the ground. The fracture toughness of the glass is
0.83 MPa-m0.5. Should he do the stunt? Show all assumptions and calculations in support of
your answer.
Given:
Fracture toughness
Kc  0.83 MPa m
Glass dimensions
Total crack length
L  3000 mm
W  100  mm
lcrack  16.2 mm
Assumptions: Weight of stuntman
Desired safety factor
Solution:
1.
2.
0.5
Weight  900  N
NFMd  3
See Mathcad file P0553.
Calculate the nominal stress based on the assumed weight of the stuntman and the glass dimensions.
Cross-section area
A  W  t
Nominal stress
σnom 
A
2
σnom  7.087 MPa
a  0.5 l crack
a  8.1 mm
Glass half-width
b  0.5 W
b  50 mm
sec 
π a 

 2 b 
β  1.017
Using equation 5.14b, calculate the stress intensity factor for the given assumptions.
0.5
K  1.149 MPa m
Using equation 5.15, calculate the safety factor against sudden failure for the given assumptions.
NFM 
5.
Weight
Crack half-width
K  β  σnom π a
4.
A  127 mm
Determine the value of the geometry factor  from equation 5.14c for a plate with a central crack.
β 
3.
t  1.27 mm
Kc
K
NFM  0.72
The stuntman should definitely not do the stunt since the factor of safety is not only less than the desired
value, but is less than one.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-54-1
PROBLEM 5-54
_____
Statement:
A material has a fracture toughness of 50 MPa-m0.5 and a yield strength of 1000 MPa and is to be
made into a large panel. If the panel is stressed to one-half the yield stress, what is the maximum
central crack size that can be tolerated without catastrophic failure?
Given:
Fracture toughness
Kc  50 MPa m
Yield strength
S y  1000 MPa
Solution:
1.
See Mathcad file P0554.
Calculate the nominal stress based on the yield strength and the stress level given in the problem statement.
σnom 
2.
0.5
Sy
σnom  500 MPa
2
Determine the value of the geometry factor  from the discussion in Section 5.3 for a large plate with a central
crack.
β  1
3.
Using equation 5.14b, calculate the critical crack length for this material under the given stress condition.
 Kc 
a   

π  β  σnom 
1
lcritical  2  a
2
a  3.18 mm
lcritical  6.4 mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-55-1
PROBLEM 5-55
_____
Statement:
A material that has a fracture toughness of 33 MPa-m0.5 is to be made into a large panel that is
2000 mm long by 250 mm wide and 4 mm thick. If the minimum allowable total crack length is 4
mm, what is the maximum tensile load in the long direction that can be applied without
catastrophic failure with a safety factor of 2.5?
Given:
Fracture toughness
0.5
Kc  33 MPa m
Panel dimensions
L  2000 mm
Total allow. crack length lcrack  4  mm
Safety factor
Solution:
1.
NFM  2.5
Calculate the allowable stress intensity factor using equation 5.15.
Kc
NFM
0.5
Kallow  13.2 MPa m
Determine the value of the geometry factor  from equation 5.14c for a plate with a central crack.
Crack half-width
a  0.5 l crack
a  2 mm
Panel half-width
b  0.5 W
b  125 mm
β 
3.
sec 
π a 

 2 b 
β  1.00
Using equation 5.14b, calculate the allowable nominal stress in the panel.
σallow 
4.
t  4  mm
See Mathcad file P0555.
Kallow 
2.
W  250  mm
Kallow
β  π a
σallow  166.5 MPa
Calculate the allowable load for the given conditions.
Fallow  σallow W  t
Fallow  167 kN
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-56-1
PROBLEM 5-56
_____
Statement:
Figure P5-19 shows an SAE 1020 cold-rolled steel bar fastened to a rigid ground plane with two
0.25-in-dia A8 steel dowel pins, hardened to HRC52. For P = 1500 lb and t = 0.25 in, find:
(a) The safety factor for each pin.
(b) The safety factor for direct bearing stress in each hole.
(c) The safety factor for tearout failure if h = 1 in.
Given:
Pin diameter
Applied load
Distance between pins
d  0.250  in
P  1500 lbf
a  2.0 in
Depth of section
Distance from right pin to load
Yield strength of bar
h  1.0 in
b  4.0 in
S yb  57 ksi
Thickness of bar
t  0.25 in
Yield strength of pin
S yp  225  ksi
Solution:
1.
See Mathcad file P0556.
Draw a free-body diagram and find the shear forces (reactions) on each pin.
a
b
RL
h
RR
P
Write equations 3.3b for the bar and solve for the reactions.
 F:
RL 
2.
b
a
P
π d
4
RR  P  RL
RR  4500 lbf
2
2
A  0.0491 in
Use equation 4.9 to determine the shear stress in each pin.
Left pin
Right pin
4.
RL  3000 lbf
RL  a  P b  0
Calculate the cross-section area of a pin.
A 
3.
 M:
RL  RR  P  0
τL 
τR 
RL
τL  61.1 ksi
A
RR
τR  91.7 ksi
A
(a) From equations 5.8c and 5.9b, the safety factor against failure in the pins is
Left pin
NL 
0.577  S yp
τL
NL  2.1
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
NR 
Right pin
5.
0.577  S yp
τR
5-56-2
NR  1.4
Calculate the bearing area from equation 4.10 and use it to determine the bearing stress in each hole.
Abear  d  t
Bearing area
σL 
σR 
RL
Abear
RR
Abear
2
Abear  0.0625 in
σL  48.0 ksi
σR  72.0 ksi
These are principal stresses 1.
6.
7.
(b) Calculate the safety factor for direct bearing from equation 5.8c where 2 and 3 are both zero.
Left hole
NL 
Right hole
NR 
The tearout area is
Atear  2  
S yb
σL
S yb
σR
NL  1.2
NR  0.8
h  d

  t , where (h - d)/2 is the distance from the edge of the hole to the
 2  
outside of the bar. Substitute this area in equation 4.9 for the shear area and solve for the shear strength xy.
Atear  2  

Left hole
Right hole
8.
h  d
2

  t
 
τL 
τR 
2
Atear  0.187 in
RL
Atear
RR
Atear
τL  16.00 ksi
τR  24.00 ksi
(c) From equations 5.8c and 5.9b, the safety factor against tearout failure in the holes is
Left hole
NL 
Right hole
NR 
0.577  S yb
τL
0.577  S yb
τR
NL  2.1
NR  1.4
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-57-1
PROBLEM 5-57
_____
Statement:
Figure P5-19 shows a class 50 cast iron bar fastened to a rigid ground plane with two 0.25-in-dia
A8 steel dowel pins, hardened to HRC52. For P = 1500 lb and t = 0.25 in, find:
(a) The safety factor for each pin.
(b) The safety factor for direct bearing stress in each hole.
(c) The safety factor for tearout failure if h = 1 in.
Given:
Pin diameter
Applied load
Distance between pins
d  0.250  in
P  1500 lbf
a  2.0 in
Depth of section
Distance from right pin to load
Tensile strength of bar
h  1.0 in
b  4.0 in
S utb  52 ksi
Thickness of bar
t  0.25 in
Yield strength of pin
S yp  225  ksi
Solution:
1.
See Mathcad file P0557.
Draw a free-body diagram and find the shear forces (reactions) on each pin.
a
b
RL
h
RR
P
Write equations 3.3b for the bar and solve for the reactions.
 F:
RL 
2.
b
a
P
π d
4
RR  P  RL
RR  4500 lbf
2
2
A  0.0491 in
Use equation 4.9 to determine the shear stress in each pin.
Left pin
Right pin
4.
RL  3000 lbf
RL  a  P b  0
Calculate the cross-section area of a pin.
A 
3.
 M:
RL  RR  P  0
τL 
τR 
RL
τL  61.1 ksi
A
RR
τR  91.7 ksi
A
(a) From equations 5.8c and 5.9b, the safety factor against failure in the pins is
Left pin
NL 
0.577  S yp
τL
NL  2.1
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
NR 
Right pin
5.
0.577  S yp
τR
5-57-2
NR  1.4
Calculate the bearing area from equation 4.10 and use it to determine the bearing stress in each hole.
Abear  d  t
Bearing area
σL 
σR 
RL
Abear
RR
Abear
2
Abear  0.0625 in
σL  48.0 ksi
σR  72.0 ksi
These are principal stresses 1.
6.
7.
(b) Calculate the safety factor for direct bearing from equation 5.12a where 2 and 3 are both zero.
Left hole
NL 
Right hole
NR 
The tearout area is
Atear  2  
S utb
σL
S utb
σR
NL  1.1
NR  0.7
h  d

  t , where (h - d)/2 is the distance from the edge of the hole to the
 2  
outside of the bar. Substitute this area in equation 4.9 for the shear area and solve for the shear strength xy.
Atear  2  

Left hole
Right hole
8.
h  d
2

  t
 
τL 
τR 
2
Atear  0.187 in
RL
Atear
RR
Atear
τL  16.00 ksi
τR  24.00 ksi
(c) For pure shear the Mohr circle is centered at 0,0 and has a radius equal to the shear stress. This results in 1
= . Using the Modified-Mohr failure theory and Figure 5-11, we see that we can use equation 5.12a for the
safety factor against tearout.
NL 
S utb
τL
NL  3.3
NR 
S utb
τR
NR  2.2
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-58-1
PROBLEM 5-58
_____
Statement:
Figure P5-20 shows a bracket machined from 0.5-in-thick SAE 1045 cold-rolled steel flat stock. It
is rigidly attached to a support and loaded with P = 5000 lb at point D. Find:
(a) The safety factor against static failure at point A.
(b) The safety factor against static failure at point B.
Given:
Distance from support to:
Point D
d  8  in
Depth of section h  3  in
Applied load
P  5000 lbf
Points B and C b  17 in
Thickness of section
t  0.5 in
Tensile yield strength
S y  77 ksi
Assumptions: The bracket remains flat and does not buckle (out-of-plane) under the applied load.
Solution:
1.
See Mathcad file P0558.
Calculate the cross-section area and moment of inertia at A, B, and C, which are the same.
2
A  h  t
2.
A  1.500 in
t h
I 
3
4
I  1.1250 in
12
For part (a), draw a free-body diagram of the entire bracket.
V
A
y
h
M
B
x
h
C
d
h
D
P
3.
Use the equilibrium equations 3.3a to calculate the shear force and bending moment at the support.
 F:
V  P
4.
 M:
V  P  0
V  5000 lbf
MA  P ( d )
MA  40000 in lbf
The normal stress in the bracket at point A is determined using equation 4.11b.
c  0.5 h
Distance from neutral axis to extreme fiber
Normal stress at point A
5.
P ( d )  M  0
σA  
MA c
I
c  1.500 in
σA  53.33 ksi
(a) The transverse shear stress in the bracket at point A is zero, therefore A is a principal stress. There are no
stress components in the y or z directions so this is a case of uniaxial stress. Thus, equations 5.7 reduce to
σ' 
2
σA
σ'  53.3 ksi
Use equation 5.8a to calculate the factor of safety against a static failure at point A.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
NA 
6.
Sy
5-58-2
NA  1.4
σ'
For part (b), draw a free-body diagram of the portion of the bracket that is below point B.
b
F
y
M
B
x
d
h
D
P
7.
Use the equilibrium equations 3.3a to calculate the normal force and bending moment on the section shown.
 F:
F  P
8.
 M:
F  P  0
F  5000 lbf
MB1  P ( b  0.5 h  d )
MB1  52500 in lbf
The normal stress in the bracket at point B in the y direction is a combination of uniform tension and bending
and is determined by summing equations 4.7 and 4.11b.
Normal stress at B in y direction
9.
P ( b  0.5 h  d )  M  0
σBy 
MB1 c
I

F
A
σBy  73.33 ksi
The normal stress in the bracket at point B in the x direction is bending and is determined from equation 4.11b,
using the FBD from part (a).
MB2  V  b  MA
MB2  45000 in lbf
Normal stress at B in x direction
σBx 
MB2 c
I
σBx  60.00 ksi
10. (b) The transverse shear at B due to the shear force V is zero so Bx and By are the only stress components at B.
Use equations 5.7d and 5.8a to determine the factor of safety against a static failure at B (ignoring the stress
concentration there).
σ' 
NB 
2
2
σBx  σBy  σBx σBy
Sy
σ'
σ'  67.66 ksi
NB  1.1
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P0558.xmcd
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-59-1
PROBLEM 5-59
_____
Statement:
Figure P5-20 shows a bracket machined from 1-in-thick class 60 cast iron flat stock. It is rigidly
attached to a support and loaded with P = 5000 lb at point D. Find:
(a) The safety factor against static failure at point A.
(b) The safety factor against static failure at point B.
Given:
Distance from support to:
Point D
d  8  in
Depth of section h  3  in
Applied load
P  5000 lbf
Points B and C b  17 in
Thickness of section
t  0.5 in
Ultimate tensile strength
S ut  62 ksi
S uc  187  ksi
Ultimate comp. strength
Assumptions: The bracket remains flat and does not buckle (out-of-plane) under the applied load.
Solution:
1.
See Mathcad file P0559.
Calculate the cross-section area and moment of inertia at A, B, and C, which are the same.
2
A  h  t
2.
A  1.500 in
t h
I 
3
4
I  1.1250 in
12
For part (a), draw a free-body diagram of the entire bracket.
V
A
y
h
M
B
x
h
C
d
h
D
P
3.
Use the equilibrium equations 3.3a to calculate the shear force and bending moment at the support.
 F:
V  P
4.
 M:
V  P  0
V  5000 lbf
MA  P ( d )
MA  40000 in lbf
The normal stress in the bracket at point A is determined using equation 4.11b.
c  0.5 h
Distance from neutral axis to extreme fiber
Normal stress at point A
5.
P ( d )  M  0
σA  
MA c
I
c  1.500 in
σA  53.33 ksi
(a) The transverse shear stress in the bracket at point A is zero, therefore A is a principal stress. There are no
stress components in the y or z directions so this is a case of uniaxial stress. Thus, use equation 5.12a (adapted
to a compressive stress state) to calculate the factor of safety against a static failure at point A.
NA 
S uc
σA
NA  3.5
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6.
5-59-2
For part (b), draw a free-body diagram of the portion of the bracket that is below point B.
b
F
y
M
B
x
d
h
D
P
7.
Use the equilibrium equations 3.3a to calculate the normal force and bending moment on the section shown.
 F:
F  P
8.
 M:
F  P  0
F  5000 lbf
MB1  P ( b  0.5 h  d )
MB1  52500 in lbf
The normal stress in the bracket at point B in the y direction is a combination of uniform tension and bending
and is determined by summing equations 4.7 and 4.11b.
σBy 
Normal stress at B in y direction
9.
P ( b  0.5 h  d )  M  0
MB1 c
I

F
A
σBy  73.33 ksi
The normal stress in the bracket at point B in the x direction is bending and is determined from equation 4.11b,
using the FBD from part (a).
MB2  V  b  MA
MB2  45000 in lbf
Normal stress at B in x direction
σBx 
MB2 c
I
σBx  60.00 ksi
10. (b) The transverse shear at B due to the shear force V is zero so Bx and By are the only stress components at B.
Use equations 4.6 to determine the principal stresses and 5.12a to determine the factor of safety against a static
failure at B (ignoring the stress concentration there).
σ1 
σ2 
σBx  σBy
2
σBx  σBy
2
σ3  0  ksi
2
 σBx  σBy 
 

2


σ1  73.333 ksi
2
 σBx  σBy 
 

2


σ2  60.000 ksi
NB 
S ut
σ1
NB  0.85
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-60-1
PROBLEM 5-60
_____
Statement:
Figure P5-21 shows a 1-in-dia SAE 1040 hot-rolled, normalized steel bar supported and subjected
to the applied load P = 500 lb. Find the safety factor against static failure.
Given:
Diameter
Applied load
d  1.00 in
P  500  lbf
Dimensions:
a  20 in
Solution:
1.
6
Modulus of elasticity
E  30 10  psi
Yield strength
S y  54 ksi
L  40 in
See Mathcad file P0560.
Draw a free-body diagram.
L
a
R2
M1
R1
2.
This is a statically indeterminate beam because there are three unknown reactions, R1, M1, and R2. To solve
for these unknowns, follow the method presented in Example 4-7. First, calculate the moment of inertia and
distance to the extreme fiber for the round section.
I 
3.
P
π d
4
4
I  0.0491 in
64
c  0.5 d
c  0.500 in
From inspection of the FBD, write the load function equation
q(x) = -M1<x>-2 + R1<x>-1 - R2<x - a>-1 + P<x - L>-1
4.
Integrate this equation from - to x to obtain shear, V(x)
V(x) = -M1<x>-1 + R1<x>0 - R2<x - a>0 + P<x - L>0
5.
Integrate this equation from - to x to obtain moment, M(x)
M(x) = -M1<x>0 + R1<x>1 - R2<x - a>1 + P<x - L>1
6.
Integrate the moment function, multiplying by 1/EI, to get the slope.
(x) = [ -M1<x>1 + R1<x>2/2 - R2<x - a>2/2 + P<x - L>2/2 + C3]/EI
5.
Integrate again to get the deflection.
y(x) = [-M1<x>2/2 + R1<x>3/6 - R2<x - a>3/6 + P<x - L>3/6 + C3x + C4]/EI
7.
Evaluate R1, M1, R2, C3 and C4
At x = 0, y = 0 and  = 0, therefore, C3 = 0 and C4 = 0.
At x = a, y = 0
At x = L+, V = M = 0
Guess
M1  1000 in lbf
Given
y(a) = 0:
V(L) = 0:
R1  500  lbf

M1
2
2
a 
R1
6
3
R2  1000 lbf
3
 a = 0  lbf  in
R1  R2  P = 0  lbf
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-60-2
M1  R1 L  R2 ( L  a ) = 0  lbf  in
M(L) = 0:
 M1 
 
 R1   Find  M1 R1 R2
R 
 2
M1  5000 in lbf
R1  750 lbf
R2  1250 lbf
x  0  in 0.02 L  L
8.
Define the range for x
9.
For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
S ( x z)  if ( x  z 1 0 )
10. Write the moment equation in Mathcad form, using the function S as a multiplying factor to get the effect of the
singularity functions.
M ( x)  M1  R1 S ( x 0  in)  x  R2 S ( x a )  ( x  a )  P S ( x L)  ( x  L)
11. Plot the moment equation and determine the maximum bending moment.
MOMENT DIAGRAM
As expected, the maximum bending
moment occurs under the support at
x = a.
Mmax  M ( a )
10
Mmax  10.0 kip in
5
M ( x)
kip  in
0
5
0
10
20
30
40
x
in
12. Use equation 4.11b to calculate the maximum bending stress in the bar.
σmax 
Mmax c
I
σmax  101.9 ksi
13. There are no other stress components present (the transverse shear is zero at the extreme fiber) so this is a
principal stress and the other two principal stresses are zero. Thus, this is a case of uniaxial stress. Determine
the safety factor against static failure using equations 5.7c and 5.8a, which reduce to
N 
Sy
σmax
N  0.53
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-61-1
PROBLEM 5-61
_____
Statement:
Figure P5-21 shows a 1.5-in-dia class 60 cast iron bar supported and subjected to the applied load
= 500 lb. Find the safety factor against static failure.
Given:
Diameter
Applied load
d  1.50 in
P  500  lbf
Dimensions:
a  20 in
Solution:
1.
6
Modulus of elasticity
E  30 10  psi
Tensile strength
S ut  54 ksi
L  40 in
See Mathcad file P0561.
Draw a free-body diagram.
L
a
R2
M1
R1
2.
This is a statically indeterminate beam because there are three unknown reactions, R1, M1, and R2. To solve
for these unknowns, follow the method presented in Example 4-7. First, calculate the moment of inertia and
distance to the extreme fiber for the round section.
I 
3.
P
π d
4
4
I  0.2485 in
64
c  0.5 d
c  0.750 in
From inspection of the FBD, write the load function equation
q(x) = -M1<x>-2 + R1<x>-1 - R2<x - a>-1 + P<x - L>-1
4.
Integrate this equation from - to x to obtain shear, V(x)
V(x) = -M1<x>-1 + R1<x>0 - R2<x - a>0 + P<x - L>0
5.
Integrate this equation from - to x to obtain moment, M(x)
M(x) = -M1<x>0 + R1<x>1 - R2<x - a>1 + P<x - L>1
6.
Integrate the moment function, multiplying by 1/EI, to get the slope.
(x) = [ -M1<x>1 + R1<x>2/2 - R2<x - a>2/2 + P<x - L>2/2 + C3]/EI
5.
Integrate again to get the deflection.
y(x) = [-M1<x>2/2 + R1<x>3/6 - R2<x - a>3/6 + P<x - L>3/6 + C3x + C4]/EI
7.
Evaluate R1, M1, R2, C3 and C4
At x = 0, y = 0 and  = 0, therefore, C3 = 0 and C4 = 0.
At x = a, y = 0
At x = L+, V = M = 0
Guess
M1  1000 in lbf
Given
y(a) = 0:
V(L) = 0:
R1  500  lbf

M1
2
2
a 
R1
6
3
R2  1000 lbf
3
 a = 0  lbf  in
R1  R2  P = 0  lbf
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-61-2
M1  R1 L  R2 ( L  a ) = 0  lbf  in
M(L) = 0:
 M1 
 
 R1   Find  M1 R1 R2
R 
 2
M1  5000 in lbf
R1  750 lbf
R2  1250 lbf
x  0  in 0.02 L  L
8.
Define the range for x
9.
For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
S ( x z)  if ( x  z 1 0 )
10. Write the moment equation in Mathcad form, using the function S as a multiplying factor to get the effect of the
singularity functions.
M ( x)  M1  R1 S ( x 0  in)  x  R2 S ( x a )  ( x  a )  P S ( x L)  ( x  L)
11. Plot the moment equation and determine the maximum bending moment.
MOMENT DIAGRAM
As expected, the maximum bending
moment occurs under the support at
x = a.
Mmax  M ( a )
10
Mmax  10.0 kip in
5
M ( x)
kip  in
0
5
0
10
20
30
40
x
in
12. Use equation 4.11b to calculate the maximum bending stress in the bar.
σmax 
Mmax c
I
σmax  30.2 ksi
13. There are no other stress components present (the transverse shear is zero at the extreme fiber) so this is a
principal stress and the other two principal stresses are zero. Thus, this is a case of uniaxial stress. Determine
the safety factor against static failure using equation 5.12a.
N 
S ut
σmax
N  1.8
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-62-1
PROBLEM 5-62
_____
Statement:
Figure P5-22 shows a pivot pin that is press-fit into part A and is slip fit in part B. If F = 100 lb,
l = 2 in, and d = 0.5 in, what is the pin's safety factor against yielding when made of SAE 1020
cold-rolled steel?
Given:
Applied force
F  100  lbf
Yield strength
S y  57 ksi
Total length, l
Pin dia
l  2.00 in
d  0.5 in
Beam length
L  0.5 l
L  1.000  in
Assumptions: 1. Since there is a slip fit between the pin and part B, part B offers no resistance to bending of
the pin and, since the pin is press-fit into part A, it can be modeled as a cantilever beam of
length l/2.
2. Part B distributes the concentrated force F so that, at the pin, it is uniformly distributed over
the exposed length of the pin.
Solution:
1.
See Mathcad file P0562.
Calculate the intensity of the uniformly distributed load acting over the length of the pin.
w 
2.
F
w  100.0 
L
lbf
in
A cantilever beam with uniform loading is shown in Figure B-1(b) in Appendix B. In this case, the dimension a
the figure is zero. As shown in the figure, when a = 0, the maximum bending moment occurs at the support and
is
2
w L
Mmax 
3.
2
Calculate the moment of inertia and distance to the extreme fiber of the pin. The bending stress in the beam is
then found using equation 4.11b.
I 
π d
4
I  3.068  10
64
c  0.5 d
σ 
4.
Mmax  50.00  lbf  in
Mmax c
I
3
4
 in
c  0.250  in
σ  4074 psi
There are no other stress components present (the transverse shear is zero at the extreme fiber) so this is a
principal stress and the other two principal stresses are zero. Thus, this is a case of uniaxial stress. Determine
the safety factor against static failure using equations 5.7c and 5.8a, which reduce to
N 
Sy
σ
N  14
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-63-1
PROBLEM 5-63
_____
Statement:
Figure P5-22 shows a pivot pin that is press-fit into part A and is slip fit in part B. If F = 100 N, l =
50 mm, and d = 16 mm, what is the pin's safety factor against yielding when made of class 50 cast
iron?
Given:
Applied force
F  100  N
Tensile strength
S ut  359  MPa
Total length, l
Pin dia
l  50 mm
d  16 mm
Beam length
L  0.5 l
L  25 mm
Assumptions: 1. Since there is a slip fit between the pin and part B, part B offers no resistance to bending of
the pin and, since the pin is press-fit into part A, it can be modeled as a cantilever beam of length
l/2.
2. Part B distributes the concentrated force F so that, at the pin, it is uniformly distributed over
the exposed length of the pin.
Solution:
1.
See Mathcad file P0563.
Calculate the intensity of the uniformly distributed load acting over the length of the pin.
w 
2.
F
w  4.0
L
N
mm
A cantilever beam with uniform loading is shown in Figure B-1(b) in Appendix B. In this case, the dimension a
the figure is zero. As shown in the figure, when a = 0, the maximum bending moment occurs at the support and
is
2
w L
Mmax 
3.
2
Calculate the moment of inertia and distance to the extreme fiber of the pin. The bending stress in the beam is
then found using equation 4.11b.
I 
π d
4
σ 
3
I  3.217  10  mm
64
c  0.5 d
4.
Mmax  1250 N  mm
Mmax c
I
4
c  8.000  mm
σ  3.108  MPa
There are no other stress components present (the transverse shear is zero at the extreme fiber) so this is a
principal stress and the other two principal stresses are zero. Thus, this is a case of uniaxial stress. Determine
the safety factor against static failure using equation 5.12a.
N 
S ut
σ
N  115
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-64-1
PROBLEM 5-64
Statement:
A differential element is subjected to the stresses (in ksi): x = 10, y = -20, xy = -20. The material
is uneven and has strengths (in ksi) of S ut = 50, S y = 40, and S uc = 90. Calculate the safety factor
and draw a a-b diagram showing the boundary for each theory with the stress state and load
line using:
(a) Coulomb-Mohr theory, and
(b) Modified Mohr theory.
Given:
Stress components
σx  10 ksi
σy  20 ksi
τxy  20 ksi
Material properties
S ut  50 ksi
S y  40 ksi
S uc  90 ksi
Solution:
1.
See Figure 5-62 and Mathcad file P0564.
Calculate the nonzero principal stresses using equation 4.6a.
σa 
σb 
2.
4.
2
σx  σy
2
2
 σx  σy 
2
 
  τxy
2


σa  20 ksi
2
 σx  σy 
2
 
  τxy
2


σb  30 ksi
Calculate the slope of load line. (The load line is the line from the origin through the stress point.)
m 
3.
σx  σy
σb
σa
m  1.5
The safety factor equation for both theories is different for each quadrant the load line falls in. The equation
for the modified Mohr factor of safety is different for each of the two regions in the 4th quadrant that the load
line can fall in. In this case, the load line falls in the 4th quadrant, below the -1 slope line.. The factors of
safety are:
(a) Coulomb-Mohr theory
Na 
(b) Modified Mohr theory
Nb 
S uc S ut
S uc σa  S ut σb
S uc
 S uc  S ut 
 S
  σa  σb
ut


Na  1.4
Nb  2
Plot the a-b diagram showing the safe-fail boundaries, the stress state point (20 ksi, -30 ksi) and the load line.
Note that if a > b , then only that area on the graph that is to the right of and below the diagonal line can
contain valid stress points. The factor of safety is the distance along the load line from the origin to the
intersection of the load line with the failure boundary, divided by the distance from the origin to the stress poin
Since the distance from the origin to the modified Mohr boundary is greater than the distance to the
Coulomb-Mohr boundary, its factor of safety is greater. See Figure 5-62 on the following page.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-64-2
b
50
40
MINIMUM NONZERO PRINCIPAL STRESS, KSI
30
20
(a) Coulomb-Mohr
boundary
10
a
0
-10
-20
(20,-30)
-30
-40
Stress states at
which failure
will occur
-50
-60
-S
(b) Modified Mohr
boundary
-70
Load Line
-80
-90
-S
ut
uc
-100
-100 -90 -80 -70 -60 -50 -40 -30 -20 -10
0
10
20
30
40
50
MAXIMUM PRINCIPAL STRESS, KSI
FIGURE 5-64
a -  b Diagram for Problem 5-64
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-65-1
PROBLEM 5-65
Statement:
A differential element is subjected to the stresses (in ksi): x = 10, y = -5, xy = 15. The material is
uneven and has strengths (in ksi) of S ut = 50, S y = 40, and S uc = 90. Calculate the safety factor and
draw a a-b diagram showing the boundary for each theory with the stress state and load line
using:
(a) Coulomb-Mohr theory, and
(b) Modified Mohr theory.
Given:
Stress components
σx  10 ksi
σy  5  ksi
τxy  15 ksi
Material properties
S ut  50 ksi
S y  40 ksi
S uc  90 ksi
Solution:
1.
See Figure 5-65 and Mathcad file P0565.
Calculate the nonzero principal stresses using equation 4.6a.
σa 
σb 
2.
4.
2
σx  σy
2
2
 σx  σy 
2
 
  τxy
 2 
σa  19.3 ksi
2
 σx  σy 
2
 
  τxy
 2 
σb  14.3 ksi
Calculate the slope of load line. (The load line is the line from the origin through the stress point.)
m 
3.
σx  σy
σb
σa
m  0.741
The safety factor equation for both theories is different for each quadrant the load line falls in. The equation
for the modified Mohr factor of safety is different for each of the two regions in the 4th quadrant that the load
line can fall in. In this case, the load line falls in the 4th quadrant, above the -1 slope line.. The factors of
safety are:
(a) Coulomb-Mohr theory
Na 
(b) Modified Mohr theory
Nb 
S uc S ut
S uc σa  S ut σb
S ut
σa
Na  1.8
Nb  2.6
Plot the a-b diagram showing the safe-fail boundaries, the stress state point (19.3 ksi, -14.3 ksi) and the load
line. Note that if a > b , then only that area on the graph that is to the right of and below the diagonal line can
contain valid stress points. The factor of safety is the distance along the load line from the origin to the
intersection of the load line with the failure boundary, divided by the distance from the origin to the stress poin
Since the distance from the origin to the modified Mohr boundary is greater than the distance to the
Coulomb-Mohr boundary, its factor of safety is greater. See Figure 5-63 on the following page.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-65-2
b
50
40
MINIMUM NONZERO PRINCIPAL STRESS, KSI
30
20
(a) Coulomb-Mohr
boundary
10
a
0
(19.3,-14.3)
-10
-20
-30
-40
Load Line
Stress states at
which failure
will occur
-50
-60
-S
ut
(b) Modified Mohr
boundary
-70
-80
-90
-S
uc
-100
-100 -90 -80 -70 -60 -50 -40 -30 -20 -10
0
10
20
30
40
50
MAXIMUM PRINCIPAL STRESS, KSI
FIGURE 5-65
a -  b Diagram for Problem 5-65
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-66-1
PROBLEM 5-66
Statement:
A differential element is subjected to the stresses (in ksi): x = -20, y = -15, xy = 15. The material
is uneven and has strengths (in ksi) of S ut = 50, S y = 40, and S uc = 90. Calculate the safety factor
and draw a a-b diagram showing the boundary for each theory with the stress state and load
line using:
(a) Coulomb-Mohr theory, and
(b) Modified Mohr theory.
Given:
Stress components
σx  20 ksi
σy  15 ksi
τxy  15 ksi
Material properties
S ut  50 ksi
S y  40 ksi
S uc  90 ksi
Solution:
1.
See Figure 5-66 and Mathcad file P0566.
Calculate the nonzero principal stresses using equation 4.6a.
σa 
σb 
2.
4.
2
σx  σy
2
2
 σx  σy 
2
 
  τxy
 2 
σa  2.29 ksi
2
 σx  σy 
2
 
  τxy
 2 
σb  32.7 ksi
Calculate the slope of load line. (The load line is the line from the origin through the stress point.)
m 
3.
σx  σy
σb
σa
m  14.263
(third quadrant since both principal
stresses are negative)
The safety factor equation for both theories is the same when the load line falls in the third quadrant. The
factors of safety are:
(a) Coulomb-Mohr theory
Na  
(b) Modified Mohr theory
Nb  
S uc
σb
S uc
σb
Na  2.8
Nb  2.8
Plot the a-b diagram showing the safe-fail boundaries, the stress state point (-2.29 ksi, -32.7 ksi) and the load
line. Note that if a > b , then only that area on the graph that is to the right of and below the diagonal line
can contain valid stress points. The factor of safety is the distance along the load line from the origin to the
intersection of the load line with the failure boundary, divided by the distance from the origin to the stress
point. Since the distance from the origin to the modified Mohr boundary is the same as the distance to the
Coulomb-Mohr boundary, its factor of safety is the same. See Figure 5-63 on the following page.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-66-2
b
50
40
MINIMUM NONZERO PRINCIPAL STRESS, KSI
30
20
(a) Coulomb-Mohr
boundary
10
a
0
-10
-20
-30
(-2.29,-32.7)
-40
Stress state at
which failure
will occur for
both theories
-50
-60
-S
ut
-70
-80
-90
-S
(b) Modified Mohr
boundary
uc
Load Line
-100
-100 -90 -80 -70 -60 -50 -40 -30 -20 -10
0
10
20
30
40
50
MAXIMUM PRINCIPAL STRESS, KSI
FIGURE 5-66
a -  b Diagram for Problem 5-66
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-67-1
PROBLEM 5-67
_____
Statement:
Derive the von Mises effective stress equation 5.7d for the two-dimensional case.
Solution:
See Mathcad file P0567.
1.
Start with equation 5.7c, which gives the von Mises stress in terms of the two nonzero principal stresses.
σ' 
2.
σ3 
(a)
σx  σy
2
σx  σy
2
2
 σx  σy 
2
 
  τxy
 2 
2
 σx  σy 
2
 
  τxy
2


σx  σy
and
2
6.
(c)
2
 σx  σy 
2
R  
  τxy
 2 
(d)
σ3  σc  R
(e)
Substitute equations d into b and c.
σ1  σc  R
5.
(b)
To make the manipulations easier, define:
σc 
4.
2
Define 1 and 3 in terms of x, y, and xy using equations 4.6a.
σ1 
3.
2
σ1  σ1 σ3  σ3
Substitute equations e into a, expand, collect terms and simplify.
σ' 
σc  R 2  σc  R  σc  R  σc  R 2
σ' 
σc  2  R σc  R  σc  R  σc  2  R σc  R
σ' 
σc  3  R
2
2
2
2
2
2
2
(f)
2
Substitute equations d into f, expand, collect terms and simplify to obtain the derived equation.
2
2
 σx  σy 
 σx  σy 
2
σ'  
  3 
  3  τxy
2
2




σ' 
2
2
σx  σy  σx σy  3  τxy
2
(5.7d)
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-68-1
PROBLEM 5-68
Statement:
Figure P5-23 shows an oil-field pump jack. The crank drive shaft at O2 is loaded in torsion and
bending with maximum values of 6500 in-lb and 9800 in-lb, respectively. The point on the shaft
with maximum stress is located away from the key that connects the shaft to the crank. Using a
factor of safety of 2 against static yielding, determine a suitable diameter for the shaft if it is to be
made of SAE 1040 cold-rolled steel.
Given:
Yield strength
S y  71 ksi
Torque T  6500 in lbf
Solution:
1.
Bending moment M  9800 in lbf
See Figure P5-23 and Mathcad file P0568.
Express the torsional and bending stresses as a functions of the unknown shaft diameter, d
Bending stress
σx( d ) 
32 M
π d
Torsional stress
3
16 T
τxy( d ) 
π d
2.
3
Use these two stresses in an expression for the von Mises effective stress, equation 5.7d withy = 0.
σ'( d ) 
von Mises effective stress
3.
Ns  2
Factor of safty
2
σx( d )  3  τxy( d )
2
Use equation 5.8a as a design relationship to solve for the diameter, d.
Design equation
2
2
σx( d )  3  τxy( d ) =
2
Sy
Ns
2
 32 M   3  16 T  =  S y 
 3
 3   Ns 
 π d 
 π d   
2
Solving for d
1
 ( 32 M ) 2  3  ( 16 T ) 2
d  

2


2  Sy 
π  



 Ns 

6
A suitable diameter for the given design requirements is
d  1.480 in
d  1.500  in
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-69-1
PROBLEM 5-69
Statement:
Figure P5-24a shows a C-clamp with an elliptical body dimensioned as shown. The clamp has a
T-section with a uniform thickness of 3.2 mm at the throat as shown in Figure P5-24b. Find the
static factor of safety if the clamping force is 2.7 kN and the material is class 40 gray cast iron.
Given:
Clamping force F  2.7 kN
Distance from center of screw to throat
Section dimensions:
t  3.2 mm
Material properties
Solution:
1.
ri  63.5 mm
Flange b  28.4 mm
Web h  31.8 mm
S ut  290  MPa
S uc  965  MPa
See Figure P4-26 and Mathcad file P0569.
Determine the location of the CG of the T-section and the distance from the centerline of the screw to the
centroid of the section at the throat.
yCG 
0.5 t ( b  t)  0.5 ( h  t)  ( h  t) t
yCG  9.578  mm
b t  ( h  t)  t
rc  ri  yCG
2.
rc  73.078 mm
Using equation 4.12a and Figure 4-16, calculate the distance to the neutral axis, rn, and the distance from the
centroidal axis to the neutral axis, e.
Distance from the screw centerline to the outside fiber
Cross section area
A  b  t  ( h  t)  t
ri t



r
i
6.
rn  71.864 mm

t
dr  
dr
r
r

r  t
i
e  rc  rn
e  1.214  mm
M  rc F
M  197  N  m
Calculate the distances from the neutral axis to the inner and outer fibers.
ci  rn  ri
5.
2
Take a section through the throat area and draw a FBD. There will be a vertical axial force through the section
CG (at a distance rc from the screw centerline) which will form a couple of magnitude rc x F. This couple will be
balanced by an internal moment of equal magnitude.
Internal moment
4.
ro
b
Distance from centroidal to neutral axis
3.
ro  95.3 mm
A  182.4  mm
A
rn 
Distance to neutral axis
ro  ri  h
ci  8.364  mm
co  ro  rn
co  23.436 mm
Using equations 4.12d and 4.12e, calculate the stresses at the inner and outer fibers of the throat section.
 ci  F

e A  ri  A
M
Inner radius
σi 
Outer radius
σo  

 co  F

e A  ro  A
M

σi  132.2  MPa
σo  204.3  MPa
These are the principal stresses so,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
7.
5-69-2
Inner radius
σ1i  σi
σ2i  0  MPa
σ3i  0  MPa
Outer radius
σ1o  0  MPa
σ2o  0  MPa
σ3o  σo
Calculate the factor of safety using equations 5.12c, 5.12d, and 5.12e.
Inner radius
C1i 
1
C2i 
1
C3i 
1
2
2
2

S uc  2  S ut

S uc

S uc  2  S ut

S uc

S uc  2  S ut

S uc
  σ1i  σ2i 
  σ2i  σ3i 
  σ3i  σ1i 

  σ1i  σ2i
C1i  92.46  MPa


  σ2i  σ3i
C2i  0.00 MPa


  σ3i  σ1i
C3i  92.46  MPa

σeff  max C1i C2i C3i σ1i σ2i σ3i
Ni 
S ut
σeff  132.182  MPa
Ni  2.2
σeff
Outer radius
C1o 
1
C2o 
1
C3o 
1
2
2
2

S uc  2  S ut

S uc

S uc  2  S ut

S uc

S uc  2  S ut

S uc
  σ1o  σ2o 
  σ2o  σ3o 
  σ3o  σ1o 

  σ1o  σ2o

  σ2o  σ3o
S ut
σeff
C2o  61.41  MPa


  σ3o  σ1o
C3o  61.41  MPa

σeff  max C1o C2o C3o σ1o σ2o σ3o
No 
C1o  0.00 MPa

σeff  61.41  MPa
No  4.7
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-70-1
PROBLEM 5-70
Statement:
A C-clamp as shown in Figure P5-24a has a rectangular cross section as in Figure P5-24c. Find the
static factor of safety if the clamping force is 1.6 kN and the material is class 50 gray cast iron.
Given:
Clamping force F  1.6 kN
Distance from center of screw to throat
Section dimensions:
Material properties
Solution:
1.
Width b  6.4 mm
S ut  359  MPa
Depth h  31.8 mm
S uc  1131 MPa
See Figure P5-24 and Mathcad file P0570.
Determine the distance from the centerline of the screw to the centroid of the section at the throat.
rc  ri 
2.
ri  63.5 mm
h
rc  79.4 mm
2
Using equation 4.12a and Figure 4-16, calculate the distance to the neutral axis, rn, and the distance from the
centroidal axis to the neutral axis, e.
Distance from the screw centerline to the outside fiber
Cross section area
Distance to neutral axis
A  b  h
rn 
A
ro
rn  78.327 mm
dr
i
e  rc  rn
e  1.073 mm
M  rc F
M  127 N  m
Calculate the distances from the neutral axis to the inner and outer fibers.
ci  rn  ri
ci  14.827 mm
co  ro  rn
co  16.973 mm
Using equations 4.12d and 4.12e, calculate the stresses at the inner and outer fibers of the throat section.
σi 
 ci  F

e A  ri  A
M
σo  
6.
2
Take a section through the throat area and draw a FBD. There will be a vertical axial force through the section
CG (at a distance rc from the screw centerline) which will form a couple of magnitude rc x F. This couple will be
balanced by an internal moment of equal magnitude.
Internal moment
5.
b
r
Distance from centroidal to neutral axis
4.
ro  95.300 mm
A  203.520 mm



r
3.
ro  ri  h

σi  143.7 MPa
 co  F

e A  ro  A
M

σo  95.8 MPa
These are the principal stresses so,
Inner radius
σ1i  σi
σ2i  0  MPa
σ3i  0  MPa
Outer radius
σ1o  0  MPa
σ2o  0  MPa
σ3o  σo
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
7.
5-70-2
Calculate the factor of safety using equations 5.12c, 5.12d, and 5.12e.
Inner radius
C1i 
1
C2i 
1
C3i 
1
2
2
2

S uc  2  S ut

S uc

S uc  2  S ut

S uc

S uc  2  S ut

S uc
  σ1i  σ2i 
  σ2i  σ3i 
  σ3i  σ1i 

  σ1i  σ2i
C1i  98.09 MPa


  σ2i  σ3i
C2i  0.00 MPa


  σ3i  σ1i
C3i  98.09 MPa

σeff  max C1i C2i C3i σ1i σ2i σ3i
Ni 
S ut
σeff  143.707 MPa
Ni  2.5
σeff
Outer radius
C1o 
1
C2o 
1
C3o 
1
2
2
2

S uc  2  S ut

S uc

S uc  2  S ut

S uc

S uc  2  S ut

S uc
  σ1o  σ2o 
  σ2o  σ3o 
  σ3o  σ1o 

  σ1o  σ2o


  σ2o  σ3o


  σ3o  σ1o

σeff  max C1o C2o C3o σ1o σ2o σ3o
No 
S ut
σeff
C1o  0.00 MPa
C2o  30.39 MPa
C3o  30.39 MPa
σeff  30.394 MPa
No  11.8
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-71-1
PROBLEM 5-71
Statement:
A C-clamp as shown in Figure P5-24a has an elliptical cross section as in Figure P5-24d.
Dimensions of the major and minor axes of the ellipse are given. Find the static factor of safety if
the clamping force is 1.6 kN and the material is class 60 gray cast iron.
Given:
Clamping force F  1.6 kN
Distance from center of screw to throat
Section dimensions:
Material properties
Solution:
1.
Width b  9.6 mm
S ut  427  MPa
Depth h  31.8 mm
S uc  1289 MPa
See Figure P5-24 and Mathcad file P0571.
Determine the distance from the centerline of the screw to the centroid of the section at the throat.
rc  ri 
2.
ri  63.5 mm
h
rc  79.4 mm
2
Using equation 4.12a and Figure 4-16, calculate the distance to the neutral axis, rn, and the distance from the
centroidal axis to the neutral axis, e.
Distance from the screw centerline to the outside fiber
Cross section area
Distance to neutral axis
b h
A  π 
2 2
ro
2

 1   r  rc 
2 b
4
2 
h


dr
i
e  rc  rn
e  0.805 mm
M  rc F
M  127 N  m
Calculate the distances from the neutral axis to the inner and outer fibers.
ci  rn  ri
ci  15.095 mm
co  ro  rn
co  16.705 mm
Using equations 4.12d and 4.12e, calculate the stresses at the inner and outer fibers of the throat section.
σi 
 ci  F

e A  ri  A
M
σo  
6.
0.5
Take a section through the throat area and draw a FBD. There will be a vertical axial force through the section
CG (at a distance rc from the screw centerline) which will form a couple of magnitude rc x F. This couple will be
balanced by an internal moment of equal magnitude.
Internal moment
5.
2
rn  78.595 mm
r
Distance from centroidal to neutral axis
4.
ro  95.300 mm
A  239.766 mm
A
rn 






r
3.
ro  ri  h

 co  F

e A  ro  A
M

σi  163.2 MPa
σo  108.7 MPa
These are the principal stresses so,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
7.
5-71-2
Inner radius
σ1i  σi
σ2i  0  MPa
σ3i  0  MPa
Outer radius
σ1o  0  MPa
σ2o  0  MPa
σ3o  σo
Calculate the factor of safety using equations 5.12c, 5.12d, and 5.12e.
Inner radius
C1i 
1
C2i 
1
C3i 
1
2
2
2

S uc  2  S ut

S uc

S uc  2  S ut

S uc

S uc  2  S ut

S uc
  σ1i  σ2i 
  σ2i  σ3i 
  σ3i  σ1i 

  σ1i  σ2i

  σ2i  σ3i
C2i  0.00 MPa


  σ3i  σ1i
C3i  109.12 MPa

σeff  max C1i C2i C3i σ1i σ2i σ3i
Ni 
C1i  109.12 MPa

S ut
σeff  163.169 MPa
Ni  2.6
σeff
Outer radius
C1o 
1
C2o 
1
C3o 
1
2
2
2

S uc  2  S ut

S uc

S uc  2  S ut

S uc

S uc  2  S ut

S uc
  σ1o  σ2o 
  σ2o  σ3o 
  σ3o  σ1o 

  σ1o  σ2o


  σ2o  σ3o


  σ3o  σ1o

σeff  max C1o C2o C3o σ1o σ2o σ3o
No 
S ut
σeff
C1o  0.00 MPa
C2o  36.02 MPa
C3o  36.02 MPa
σeff  36.016 MPa
No  11.9
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-72-1
PROBLEM 5-72
Statement:
A C-clamp as shown in Figure P5-24a has a trapezoidal cross section as in Figure P5-24e. Find the
static factor of safety if the clamping force is 350 lb and the material is class 40 gray cast iron.
Given:
Clamping force F  1.6 kN
Distance from center of screw to throat
Solution:
1.
Section dimensions:
Width b i  9.6 mm
b o  3.2 mm
Material properties
S ut  290  MPa
S uc  965  MPa
Determine the distance from the centerline of the screw to the centroid of the section at the throat.
h bi  2 bo

3 bi  bo
rc  76.75 mm
Using equation 4.12a and Figure 4-16, calculate the distance to the neutral axis, rn, and the distance from the
centroidal axis to the neutral axis, e.
Distance from the screw centerline to the outside fiber
Cross section area
Distance to neutral axis
A 
bi  bo
2
h
ro
bi 
h
rn  75.771 mm
  r  ri
dr
i
e  rc  rn
e  0.979 mm
M  rc F
M  123 N  m
Calculate the distances from the neutral axis to the inner and outer fibers.
ci  rn  ri
ci  12.271 mm
co  ro  rn
co  19.529 mm
Using equations 4.12d and 4.12e, calculate the stresses at the inner and outer fibers of the throat section.
σi 
 ci  F

e A  ri  A
M
σo  
6.
bi  bo
2
Take a section through the throat area and draw a FBD. There will be a vertical axial force through the section
CG (at a distance rc from the screw centerline) which will form a couple of magnitude rc x F. This couple will be
balanced by an internal moment of equal magnitude.
Internal moment
5.
ro  95.300 mm
A  203.520 mm
r
Distance from centroidal to neutral axis
4.
ro  ri  h
A
rn 





r
3.
Depth h  31.8 mm
See Figure P5-24 and Mathcad file P0572.
rc  ri 
2.
ri  63.5 mm

σi  126.9 MPa
 co  F

e A  ro  A
M

σo  118.4 MPa
These are the principal stresses so,
Inner radius
σ1i  σi
σ2i  0  MPa
σ3i  0  MPa
Outer radius
σ1o  0  MPa
σ2o  0  MPa
σ3o  σo
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
obtained from
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P0572.xmcd
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
7.
5-72-2
Calculate the factor of safety using equations 5.12c, 5.12d, and 5.12e.
Inner radius
C1i 
1
C2i 
1
C3i 
1
2
2
2

S uc  2  S ut

S uc

S uc  2  S ut

S uc

S uc  2  S ut

S uc
  σ1i  σ2i 
  σ2i  σ3i 
  σ3i  σ1i 

  σ1i  σ2i

  σ2i  σ3i
C2i  0.00 MPa


  σ3i  σ1i
C3i  88.77 MPa

σeff  max C1i C2i C3i σ1i σ2i σ3i
Ni 
C1i  88.77 MPa

S ut
σeff  126.907 MPa
Ni  2.3
σeff
Outer radius
C1o 
1
C2o 
1
C3o 
1
2
2
2

S uc  2  S ut

S uc

S uc  2  S ut

S uc

S uc  2  S ut

S uc
  σ1o  σ2o 
  σ2o  σ3o 
  σ3o  σ1o 

  σ1o  σ2o


  σ2o  σ3o


  σ3o  σ1o

σeff  max C1o C2o C3o σ1o σ2o σ3o
No 
S ut
σeff
C1o  0.00 MPa
C2o  35.58 MPa
C3o  35.58 MPa
σeff  35.577 MPa
No  8.2
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-73-1
PROBLEM 5-73
Statement:
The connecting rod (3) on the oil-field pump jack shown in Figure P5-23 is, in fact, made up of
two rods, one connecting on each side of the walking beam (4). Determine a suitable width of
1/2-inch-thick SAE 1020 cold-rolled bar stock to use if the maximum tensile load on the bars is
3500 lb each. Use a factor of safety of 4 against static yielding.
Given:
Yield strength
S y  57 ksi
Factor of safty
Ns  4
Thickness
t  0.50 in
Tensile force
F  3500 lbf
Solution:
1.
See Figure P5-23 and Mathcad file P0573.
Express the tensile stress as a functions of the unknown width, w.
Tensile stress
2.
w t
The tensile stress is the only stress present so it is also the von Mises effective stress.
σ'( w) 
von Mises effective stress
3.
F
σx( w) 
F
w t
Use equation 5.8a as a design relationship to solve for the diameter, d.
Design equation
Solving for w
F
w t
w 
=
Sy
Ns
N s F
t Sy
w  0.491 in
A suitable size for the given design requirements is
w  0.500  in
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-74-1
PROBLEM 5-74
Statement:
A work platform is elevated on the end of a boom that has the ability to extend its length and
vary its angle with respect to ground. The platform width is large compared to the boom
diameter so that it is possible to load the boom eccentrically resulting in a combination of
bending, torsion and direct compression in the base of the boom. At the base the boom is a
hollow tube with an outside diameter of 8 in and a wall thickness of 0.75 in. It is made from SAE
1030 CR steel. Determine the factor of safety against static failure if the loading at a point at the
base of the boom is: M = 600 kip-in, T = 76 kip-in, and an axial compression of 4800 lb.
Given:
Yield strength SAE 1030 CR steel S y  64 ksi
D  8.00 in
Boom dimensions
Loading
Solution:
1.
2.
3.
5.
F  4800 lbf
Calculate the bending stress at the point of interest.
Inside diameter
d  D  2  twall
Moment of inertia
I 
Distance to outer fiber
c  0.5 D
Bending stress
σbend 

64
π
4
 D d
d  6.500 in

4
4
I  113.438 in
c  4.000 in
M c
σbend  21.157 ksi
I
Calculate the axial stress due to the compressive load at the point of interest.

4
π
Cross-section area
A 
Axial stress
σaxial 
2
 D d

2
2
A  17.082 in
F
σaxial  0.281 ksi
A
Combine the bending and axial stresses to get the maximum normal stress on the compressive side of the boom.
σx  σbend  σaxial
σx  21.438 ksi
Calculate the torsional stress at the point of interest.
Polar moment
J  2  I
Torsional stress
τxy 
4
J  226.876 in
Tc
τxy  1.34 ksi
J
Calculate the von Mises effective stress using equation 5.7d.
von Mises stress
6.
T  76 kip in
See Mathcad file P0574.
Max. normal stress
4.
M  600  kip in
twall  0.75 in
σ' 
2
σx  3  τxy
2
σ'  21.563 ksi
Calculate the factor of safety using equation 5.8a.
Factor of safety
N 
Sy
σ'
N  2.97
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-75-1
PROBLEM 5-75
Statement:
Repeat Problem 5-74 for a boom that is made from class 20 gray cast iron. At the base the boom is
hollow tube with an outside diameter of 10 in and a wall thickness of 1.00 in.
Given:
Strength Class 20 gray cast iron
S ut  22 ksi
S uc  83 ksi
Boom dimensions
D  10.00  in
twall  1.00 in
Loading
Solution:
1.
2.
3.
4.
5.
M  600  kip in
T  76 kip in
F  4800 lbf
See Mathcad file P0575.
Calculate the bending stress at the point of interest.
Inside diameter
d  D  2  twall
Moment of inertia
I 
Distance to outer fiber
c  0.5 D
Bending stress
σbend 

64
π
4
 D d
d  8.000 in

4
4
I  289.812 in
c  5.000 in
M c
σbend  10.352 ksi
I
Calculate the axial stress due to the compressive load at the point of interest.

4
π
Cross-section area
A 
Axial stress
σaxial 
2
 D d

2
F
2
A  28.274 in
σaxial  0.17 ksi
A
Combine the bending and axial stresses to get the maximum normal stress on the tensile and compressive sides
of the boom.
Max compressive
σxc  σbend  σaxial
σxc  10.521 ksi
Max tensile
σxt  σbend  σaxial
σxt  10.182 ksi
Calculate the torsional stress at the point of interest.
Polar moment
J  2  I
Torsional stress
τxy 
4
J  579.624 in
Tc
τxy  0.656 ksi
J
Calculate the principal stresses on the tensile and compressive sides of the boom.
2
Compressive side
 σxc 
2
τmaxc  
  τxy
 2 
σ1c 
σxc
2
 τmaxc
τmaxc  5.301 ksi
σ1c  0.041 ksi
σ2c  0  ksi
σ3c 
σxc
2
 τmaxc
σ3c  10.562 ksi
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-75-2
2
 σxt 
2
τmaxt     τxy
2
 
Tensile side
σ1t 
σxt
2
τmaxt  5.133 ksi
 τmaxc
σ1t  10.392 ksi
σ2t  0  ksi
σ3t 
6.
σxt
2
 τmaxc
σ3t  0.210 ksi
Calculate the factor of safety using equations 5.12c, 5.12d, and 5.12e.
Compressive side
C1c 
1
C2c 
1
C3c 
1
2
2
2

S uc  2  S ut

S uc

S uc  2  S ut

S uc

S uc  2  S ut

S uc
  σ1c  σ2c 
  σ2c  σ3c 
  σ3c  σ1c 

  σ1c  σ2c

  σ2c  σ3c
C2c  2.80 ksi


  σ3c  σ1c
C3c  2.83 ksi

σeff  max C1c C2c C3c σ1c σ2c σ3c
Nc 
C1c  0.03 ksi

S ut
σeff  2.829 ksi
Nc  7.8
σeff
Tensile side
C1t 
1
C2t 
1
C3t 
1
2
2
2

S uc  2  S ut

S uc

S uc  2  S ut

S uc

S uc  2  S ut

S uc
  σ1t  σ2t 
  σ2t  σ3t 
  σ3t  σ1t 

  σ1t  σ2t

  σ2t  σ3t
S ut
σeff
C2t  0.06 ksi


  σ3t  σ1t
C3t  7.69 ksi

σeff  max C1t C2t C3t σ1t σ2t σ3t
Nt 
C1t  7.64 ksi

σeff  10.392 ksi
Nt  2.1
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-76-1
PROBLEM 5-76
Statement:
Assume that the curved beam of Problem 5-70 has a crack on its inside surface of half-width a =
1.5 mm and a fracture toughness of 35 MPa-m0.5. What is its safety factor against sudden
fracture?
Given:
Width of section
t  31.8 mm
Half crack length
a  1.5 mm
Solution:
Fracture toughness
See Figure 5-38 and Mathcad file P0538.
1.
From Problem 5-70, the nominal stress at the inside radius is:
Nominal inside stress
σi  143.7  MPa
2.
Calculate the half-width of the beam.
3.
Calculate the geometry and stress intensity factors.
π a 
β  sec 
β  1.006

 2 b 
K  β  σi π a
4.
Kc  35 MPa m
b  0.5 t
b  15.9 mm
K  9.92 MPa m
Determine the factor of safety against sudden fracture failure
NFM 
Kc
K
NFM  3.5
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-77-1
PROBLEM 5-77
_____
Statement:
A large aircraft panel is to be made from 7075-T651 aluminum bar. From test data it is found that
the nominal tensile stress in the panel is 200 MPa. What is the average maximum central crack
size that can be tolerated without catastrophic failure?
Given:
Fracture toughness
Kc  22 ksi in
Nominal stress
σnom  200  MPa
Solution:
1.
0.5
0.5
Kc  24.2 MPa m
See Mathcad file P0577.
Determine the value of the geometry factor  from the discussion in Section 5.3 for a large plate with a central
crack.
β  1
2.
Using equation 5.14b, calculate the critical crack length for this material under the given stress condition.
 Kc 
a   

π  β  σnom 
1
lcritical  2  a
2
a  4.65 mm
lcritical  9.3 mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-78-1
PROBLEM 5-78
Statement:
Design the connecting rod (link 3) of Problem 3-50 for a safety factor of 4 if the link is made from
SAE 1010 hot-rolled steel sheet, the pin hole diameter at each end is 6 mm, and the maximum
applied tensile load is 2000 N. There are two links carrying the load.
Given:
Force on links
Ftotal  2000 N
Yield strength
S y  179  MPa
Design safety factor
Nd  4
Pin hole diameter
d  6  mm
w  3  d
Assumptions: Choose a suitable width, say
Solution:
w  18 mm
See Figure P3-22 and Mathcad file P0577.
F 
1. The force on each link is
Ftotal
F  1000 N
2
2. With only a tensile force acting on the link, the tensile stress will be the principal stress and it will also be the
von Mises effective stress, so we have σx = 1 = '.
3. The tensile stress on each link is
F
σx =
4. Using the distortion-energy failure theory,
A
Nd =
t 
5. Solving for the thickness,t,
=
Sy
σ'
F
t w
=
= σ'
t w Sy
F
F  Nd
t  1.241  mm
w S y
t  2  mm
6. Round this up to the next higher integer value,
N 
7. The realized factor of safety
against tensile failure is,
8.
t  w S y
N  6.4
F
Check the factor of safety against bearing failure in the pin holes.
Bearing area
Abear  w t
σbear 
Abear  36.0 mm
F
Abear
2
σbear  27.8 MPa
This is the principal stresses 1.
9.
Calculate the safety factor for direct bearing from equation 5.8c where 2 and 3 are both zero.
Pin hole
10. The tearout area is
Nbear 
2
Sy
σbear
Nbear  6.4
2
Atear = 2  t R  ( 0.5 d ) , where R  0.5 w (see figure below). Substitute this area in
equation 4.9 for the shear area and solve for the shear strength xy.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-78-2
Tearout length
Shear area
2
Atear  2  t R  ( 0.5 d )
Atear  33.941 mm
Shear stress
τxy 
2
2
F
Atear
τxy  29.46  MPa
d
8.
R
From equations 5.8c and 5.9b, the safety factor against tearout failure in the holes is
Ntear 
0.577  S y
τxy
Ntear  3.5
This is slightly less than the design FS of Nd  4 so, choose t = 2.5 mm or increase w to 4*d.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-79-1
PROBLEM 5-79
Statement:
Design the compacting ram (link 4) of Problem 3-50 for a safety factor of 4 if the ram is made
from SAE 1010 hot-rolled steel bar, the pin hole diameter at the joint where link 3 attaches is 6
mm, and the applied load Fcom = 2000 N. The piston has a diameter of 35 mm.
Given:
Force at point P
Fcom  2000 N
Yield strength
S y  179  MPa
Design safety factor
Nd  4
Pin hole diameter
d  6  mm
Assumptions: The points of maximum stress on a plane through point D are sufficiently removed from point D
that there is no stress concentration at those points.
Solution:
See Figure P3-22 and Mathcad file P0579.
1. From Problem 3-51 the forces and reactions on the ram
are:
F34x  553  N
F34y  2000 N
F14E  357  N
F14F  196  N
42.5
E
F14E
D
F34x
120.0
2. The maximum bending moment is at point D and is:
M  42.5 mm F14E
F34y
M  15172.5 N  mm
F14F
The section modulus and area for the ram are
3
Z ( D) 
π D
π D
P
4
3. Between points D and P there is a compressive force of
Fcom  2000 N. Thus, there is a compressive stress due
Fcom
to this force in addition to the bending stress at point D.
Compacting Ram (4)
On the left side of the ram at the section through point D
σbL ( D) 
M
Z ( D)
F
2
A ( D) 
32
σa( D) 
77.5
Fcom
σL ( D)  σbL ( D)  σa( D)
A ( D)
On the right side of the ram at the section through point D
σbR( D)  
M
Z ( D)
σa( D) 
4  Fcom
σR( D)  σbR( D)  σa( D)
2
π D
The compressive stress on the right side will be numerically greater than that on the left side.
4. Since the shear stress due to bending is zero at these points, the axial stress will be the principal stress and it
will also be the von Mises effective stress, so we have σx = 1 = '.
5. Using the distortion-energy failure theory,
Nd =
Sy
σ'
=
Sy
σR( D)
6. Solving for the diameter, D,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
Guess
5-79-2
D  10 mm
f ( D)  Nd  σR( D)  S y
D  root( f ( D) D)
7. Round this up to the next higher even integer value, say
8. The realized factor of safety
against axial yeilding is,
N 
Sy
σR( D)
D  16.368 mm
D  18 mm
N  5.2
9. The axial stress on each side of the ram on a section through D is:
σL ( D)  18.6 MPa
σR( D)  34.4 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-80-1
PROBLEM 5-80
Statement:
A differential element is subected to the stresses given below and a ductile material has the
strengths given below. Calculate the safety factor and draw 1-3 diagrams of each theory
showing the stress state using:
(a) Maximum shear-stress theory, and
(b) Distortion-energy theory.
Given:
Principal stresses
σ1  70 MPa
Material properties
S ut  350  MPa S y  280  MPa
Solution:
σ2  0  ksi
σ3  140  MPa
S uc  350  MPa
See Figure 5-80 and Mathcad file P0580.
1. Calculate the slope of load line. (The load line is the line from the origin through the stress point.)
m 
σ3
m  2
σ1
2. The safety factor equation for the distortion-enrgy theory is the same regardless of which quadrant the load line
falls in. However, the equation for the maximum shear-stress factor of safety is different for each of the three
quadrants that the load line (1st, 3rd, or 4th) can fall in. In this case, the load line falls in the 4th quadrant. The
factors of safety are:
Na 
(a) Maximum shear-stress theory
Sy
Na  1.3
σ1  σ3
(b) Distortion energy theory
3
280
σ' 
2
σ1  σ1 σ3  σ3
(a) Maximum shear
stress boundary
2
210
(b) Distortion energy
boundary
σ'  26.9 ksi
Sy
σ'
Nb  1.5
3. Plot the 1-3 diagram showing the
safe-fail boundaries, the stress state point
(70MPa, -140 MPa) and the load line.
Note that if 1 > 3 , then only that area
on the graph that is to the right of and
below the diagonal line can contain valid
stress points. The factor of safety is the
distance along the load line from the
origin to the intersection of the load line
with the failure boundary, divided by the
distance from the origin to the stress
point. Since the distance from the origin
to the distortion-energy boundary is
greater than the distance to the maximum
shear-stress baoundary, its factor of
safety is greater.
MINIMUM NONZERO PRINCIPAL STRESS, MPa
Nb 
140
70
0
sy
1
-70
(70,-140)
-140
-210
Stress states at
which failure
will occur
-s y
-280
Load Line
-350
-420
-280
-210
-140
-70
0
70
140
210
280
350
MAXIMUM PRINCIPAL STRESS, MPa
FIGURE 5-80
1 -  3 Diagram for Problem 5-80
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-81-1
PROBLEM 5-81
Statement:
A part has the combined stress state and strengths given below. Using the Distortion-Energy
failure theory, find the von Mises effective stress and factor of safety against static failure.
Given:
Stresses: σx  70 MPa
S y  126  MPa
Strengths:
Solution:
1.
σy  35 MPa
τxy  31.5 MPa
S ut  140  MPa S uc  140  MPa
See Mathcad file P0581.
Find the maximum shear stress and principal stresses that result from this combination of applied stresses
using equations 4.6.
Maximum shear stress
2
 σx  σy 
2
τmax  
  τxy
 2 
Principal stresses
σ1 
σ2 
σx  σy
2
σx  σy
2
τmax  36.0 MPa
 τmax
σ1  88.5 MPa
 τmax
σ2  16.5 MPa
σ3  0  psi
2.
Find the von Mises effective stress using equation 5.7d:
σ' 
3.
2
2
σx  σx σy  σy  3  τxy
2
The safety factor can now be found using equation 5.8a.
σ'  81.6 MPa
N 
Sy
σ'
N  1.5
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-82-1
PROBLEM 5-82
Statement:
Repeat Problem 5-78 for the connecting rod made from class 20 cast iron.
Given:
Force on links
Ftotal  2000 N
Strength
S ut  152  MPa
Design safety factor
Nd  4
Pin hole diameter
d  6  mm
w  4  d
Assumptions: Choose a suitable width, say
Solution:
S uc  572  MPa
w  24 mm
See Figure P3-22 and Mathcad file P0582.
F 
1. The force on each link is
Ftotal
F  1000 N
2
2. With only a tensile force acting on the link, the tensile stress will be the principal stress so we have σx = 1.
3. The tensile stress on each link is
F
σx =
4. Using the modified-Mohr failure theory,
A
Nd =
t 
5. Solving for the thickness,t,
=
S ut
σ1
F
t w
= σ1
t w S ut
=
F
F  Nd
t  1.096  mm
w S ut
t  2  mm
6. Round this up to the next higher integer value,
N 
7. The realized factor of safety
against tensile failure is,
8.
t w S ut
N  7.3
F
Check the factor of safety against bearing failure in the pin holes.
Bearing area
Abear  w t
σbear 
Abear  48.0 mm
F
Abear
2
σbear  20.8 MPa
This is the principal stresses 1.
9.
Calculate the safety factor for direct bearing from equation 5.8c where 2 and 3 are both zero.
Pin hole
10. The tearout area is
Nbear 
2
S uc
σbear
Nbear  27.5
2
Atear = 2  t R  ( 0.5 d ) , where R  0.5 w (see figure below). Substitute this area in
equation 4.9 for the shear area and solve for the shear strength xy.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-82-2
Tearout length
Shear area
2
Atear  2  t R  ( 0.5 d )
Atear  46.476 mm
Shear stress
τxy 
2
2
F
Atear
τxy  21.52  MPa
Principal stress
8.
σ1  τxy
d
R
For pure shear the Mohr circle is centered at 0,0 and has a radius equal to the shear stress. This results in
1 = . Using the Modified-Mohr failure theory and Figure 5-11, we see that we can use equation 5.12a for
the safety factor against tearout.
Ntear 
S ut
σ1
Ntear  7.1
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-83-1
PROBLEM 5-83
Statement:
Repeat Problem 5-79 for the part made from class 20 cast iron.
Given:
Force at point P
Fcom  2000 N
Tensile strength
S ut  152  MPa
Design safety factor
Nd  4
Pin hole diameter
d  6  mm
Assumptions: The points of maximum stress on a plane through point D are sufficiently removed from point D
that there is no stress concentration at those points.
Solution:
See Figure P3-22 and Mathcad files P0579 and P0583.
1. From Problem 3-51 the forces and reactions on the ram
are:
F34x  553  N
F34y  2000 N
F14E  357  N
F14F  196  N
42.5
E
F14E
D
F34x
120.0
2. The maximum bending moment is at point D and is:
M  42.5 mm F14E
F34y 77.5
M  15172.5 N  mm
F14F
The section modulus and area for the ram are
3
Z ( D) 
F
2
π D
A ( D) 
32
π D
P
4
3. Between points D and P there is a compressive force of
Fcom  2000 N. Thus, there is a compressive stress due
Fcom
to this force in addition to the bending stress at point D.
Compacting Ram (4)
On the left side of the ram at the section through point D
σbL ( D) 
M
Z ( D)
σa( D) 
Fcom
σL ( D)  σbL ( D)  σa( D)
A ( D)
On the right side of the ram at the section through point D
σbR( D)  
M
Z ( D)
σa( D) 
4  Fcom
σR( D)  σbR( D)  σa( D)
2
π D
The tensile stress on the left side will be critical for an uneven, brittle material.
4. With only a tensile stress acting on the ram at this point, it will be the principal stress so we have σL = 1.
5. Using the modified-Mohr failure theory,
Nd =
S ut
σ1
=
S ut
σL
6. Solving for the diameter, D,
Guess
D  10 mm
f ( D)  Nd  σL ( D)  S ut
D  root( f ( D) D)
D  14.567 mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
7. Round this up so that
D  3  d
8. The realized factor of safety
against axial failure is,
N 
S ut
σL ( D)
5-83-2
D  18 mm
N  8.2
9. The axial stress on each side of the ram on a section through D is:
σL ( D)  18.6 MPa
σR( D)  34.4 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-84-1
PROBLEM 5-84
Statement:
A differential element is subected to the stresses and strengths given below. Calculate the
safety factor and draw 1-3 diagrams of each theory showing the stress state using:
(a) Coulomb-Mohr theory, and
(b) Modified Mohr theory.
Given:
Principal stresses
σ1  70 MPa
Material properties
S ut  350  MPa S uc  630  MPa
Solution:
σ2  0  MPa
σ3  140  MPa
See Figure 5-84 and Mathcad file P05384.
1. Calculate the slope of load line. (The load line is the line from the origin through the stress point.)
m 
σ3
σ1
m  2
2. The safety factor equation for both theories is different for each quadrant the load line falls in. The equation for
the modified Mohr factor of safety is different for each of the two regions in the 4th quadrant that the load line can
fall in. In this case, the load line falls in the 4th quadrant, below the -1 slope line.. The factors of safety are:
Na 
(a)Coulomb-Mohr theory
(b) Modified Mohr theory
S uc
280
 S uc  S ut 
 S
  σ1  σ3
ut


210
Nb  3.2
3. Plot the 1-3 diagram showing the
safe-fail boundaries, the stress state point
(70 MPa,-140 MPa) and the load line. Note
that if 1 > 3 , then only that area on the
graph that is to the right of and below the
diagonal line can contain valid stress
points. The factor of safety is the distance
along the load line from the origin to the
intersection of the load line with the failure
boundary, divided by the distance from
the origin to the stress point. Since the
distance from the origin to the modified
Mohr boundary is greater than the
distance to the Coulomb-Mohr boundary,
its factor of safety is greater.
Na  2.4
S uc σ1  S ut σ3
3
350
140
MINIMUM NONZERO PRINCIPAL STRESS, MPa
Nb 
S uc S ut
(a) Coulomb-Mohr
boundary
70
1
0
-70
(70,-140)
-140
-210
-280
Stress states at
which failure
will occur
-350
-420
-S
ut
(b) Modified Mohr
boundary
-490
Load Line
-560
-630 -S
uc
-700
-700 -630 -560 -490 -420 -350 -280 -210 -140 -70
0
70
140 210 280 350
MAXIMUM PRINCIPAL STRESS, MPa
FIGURE 5-84
1 -  3 Diagram for Problem 5-84
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5-85-1
PROBLEM 5-85
Statement:
A part has the combined stress state and strengths given below. Using the Modified-Mohr failure
theory, find the effective stress and factor of safety against static failure.
Given:
Stresses: σx  70 MPa
Strengths:
Solution:
σy  35 MPa
S y  126  MPa
τxy  31.5 MPa
S ut  140  MPa S uc  560  MPa
See Mathcad file P0585.
1.
Because S uc is greater than S ut, this is an uneven material, which is characteristic of a brittle material.
2.
Find the maximum shear stress and principal stresses that result from this combination of applied stresses using
equations 4.6.
Maximum shear stress
2
 σx  σy 
2
τmax  
  τxy
 2 
Principal stresses
σ1 
σ2 
σx  σy
2
σx  σy
2
τmax  36.0 MPa
 τmax
σ1  88.5 MPa
 τmax
σ2  16.5 MPa
σ3  0  psi
3.
4.
Find the Dowling factors C1, C2, C3 using equations 5.12b:
C1 
1
C2 
1
C3 
1
2
2
2

S uc  2  S ut

S uc

S uc  2  S ut

S uc

S uc  2  S ut

S uc
  σ1  σ2 
  σ2  σ3 
  σ3  σ1 

  σ1  σ2
C1  62.3 MPa


  σ2  σ3
C2  12.3 MPa


  σ3  σ1
C3  66.4 MPa

Then find the largest of the six stresses C1, C2, C3 , 1, 2, 3:
C
  1  
  C2  
 C 
3
σeff  max   
  σ1  
 
  σ2  
  σ3  
σeff  88.5 MPa
which is the modified-Mohr effective stress.
5.
The safety factor can now be found using equation 5.12d.
N 
S ut
σeff
N  1.6
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-1a-1
PROBLEM 6-1a
Statement:
For the data in row a in Table P6-1, find the stress range, alternating stress component, mean stress
component, stress ratio, and amplitude ratio.
Given:
σmax  1000
Solution:
See Mathcad file P0601a.
1.
σmin  0
Use equations (6.1) to calculate the required quantities.
Stress range
Δσ  σmax  σmin
Alternating stress
σa 
Mean stress
σm 
Stress ratio
R 
Amplitude ratio
A 
σmax  σmin
2
σmax  σmin
2
σmin
σmax
σa
σm
Δσ  1000
σa  500
σm  500
R0
A1
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-2a-1
PROBLEM 6-2a
Statement:
For the strength data in row a in Table P6-2, calculate the uncorrected endurance limit and draw th
strength-life (S-N) diagram for the material, assuming it to be steel.
Given:
Tensile strength
Solution:
See Mathcad file P0602a.
1.
S ut  90 ksi
Using equation (6-5a), calculate the uncorrected endurance limit.
S'e 
return 0.5 S ut if S ut  200  ksi
S'e  45.0 ksi
100  ksi otherwise
S m  0.9 S ut
S m  81.0 ksi
2.
Using equation (6.9), calculate the fatigue strength at N = 10 3 cycles.
3.
The equation for the S-N curve in the HCF region is given by equation (6.10a):
4.
Determine the constants a and b from equations (6.10c) and (6.10a). From Table 6-5, for N = 10 6 , z  3.000
b 
a 
5.
1
z
 Sm 

 S'e 
 log
b
b  0.0851
Sm
103
S'f = a  N
a  145.8 ksi
b
To draw the S-N graph over the range 10 3 <= N <= 10 8, define a piecewise continuous function.
S'f ( N ) 
return a  N
b
if N  10
6
S'e otherwise
6. Plot the S-N curve over the range
3
5
N  10 10  10
8
100
S' f ( N )
ksi
10
3
1 10
4
1 10
5
6
1 10
1 10
7
1 10
8
1 10
N
FIGURE 6-2a
S-N Diagram for Steel for Problem 6-2a
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MACHINE DESIGN - An Integrated Approach,4th Ed.
6-3-1
PROBLEM 6-3
Statement:
For the bicycle pedal-arm assembly in Figure P6-1 assume a rider-applied force that ranges from 0
1500 N at the pedal each cycle. Determine the fluctuating stresses in the 15-mm-dia pedal arm. Fi
the fatigue safety factor if S ut = 500 MPa.
Given:
Material yield strength
S y  350  MPa
S ut  500  MPa
Applied load
Fmax  1500 N
Fmin  0  N
Pedal arm diameter
d  15 mm
Solution:
1.
See Figures 6-3 and Mathcad file P0603.
From problem 4-3, the maximum principal stresses in the pedal arm due to Fmax are at point A and are
σ1max  793  MPa
2.
σ2max  0  MPa
Using equation 5.7c, the maximum von Mises stress is
2
σ'max 
σ1max  σ1max σ3max  σ3max
2
σ'max  804.7 MPa
σ'min  0  MPa
3.
The minimum von Mises stress is zero.
4.
The alternating and mean components of the von Mises stress are:
σ'a 
σ'm 
5.
σ3max  23 MPa
σ'max  σ'min
σ'a  402.4 MPa
2
σ'max  σ'min
σ'm  402.4 MPa
2
S'e  0.5 S ut
Calculate the unmodified endurance limit.
S'e  250 MPa
z
z
a
A
Tc
C
Section C
Frider
Mc
b
Arm
Arm
Fc
Pedal
y
x
y
x
FIGURE 6-3A
FIGURE 6-3B
Free Body Diagram for Problem 6-3
Points A and B at Section C
6.
B
Calculate the endurance limit modification factors for a nonrotating round beam.
Load
Cload  1
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MACHINE DESIGN - An Integrated Approach,4th Ed.
A95  0.010462 d
Size
d equiv 
6-3-2
2
A95  2.354 mm
A95
d equiv  5.544 mm
0.0766
 d equiv 
Csize  1.189  

 mm 
A  4.51
Surface
Csurf
7.
 Sut 
 A  

 MPa 
Temperature
Ctemp  1
Reliability
Creliab  0.753
 0.097
Csize  1.007
b  0.265
Csize  1
(machined)
b
Csurf  0.869
(R = 99.9%)
Calculate the modified endurance limit.
S e  Cload  Csize Csurf  Ctemp Creliab S'e
8.
2
S e  163.56 MPa
Assuming a Case 3 load line, use equation (6.18e) to determine the factor of safety.
Nf 
S e S ut
σ'a S ut  σ'm S e
Nf  0.31
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MACHINE DESIGN - An Integrated Approach
6-4a-1
PROBLEM 6-4a
Statement:
For the strength data in row a in Table P6-2, calculate the uncorrected fatigue strength at 5E8
cycles and draw the strength-life (S-N) diagram for the material, assuming it to be an aluminum
alloy.
Given:
Tensile strength
Solution:
See Mathcad file P0604a.
1.
S ut  90 ksi
Using equation (6-5c), calculate the uncorrected fatigue strength at 5E8 cycles.
S'f5E8 
return 0.4 S ut if S ut  48 ksi
S'f5E8  19.0 ksi
19 ksi otherwise
S m  0.9 S ut
S m  81.0 ksi
2.
Using equation (6.9), calculate the fatigue strength at N = 10 3 cycles.
3.
The equation for the S-N curve in the HCF region is given by equation (6.10a):
4.
Determine the constants a and b from equations (6.10c) and (6.10a). From Table 6-5, for N = 5E8 , z  5.699
b 
a 
1
z
 Sm 

 S'f5E8 
 log
b
b  0.1105
Sm
103
S'f = a  N
a  173.772 ksi
b
5.
To draw the S-N graph over the range 10 3 <= N <= 10 8,
6.
Plot the S-N curve over the range
S'f ( N )  a  N
3
5
b
8
N  10 1.01 10  10
100
S' f ( N )
ksi
10
3
1 10
4
1 10
5
6
1 10
1 10
7
1 10
8
1 10
N
FIGURE 6-4a
S-N Diagram for Aluminum for Problem 6-4a
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-5a-1
PROBLEM 6-5a
Statement:
For the data in row a in Table P6-3, find the corrected endurance strength (or limit), create
equations for the S-N line, and draw the S-N diagram.
Given:
Material
Tensile strength
steel
S ut  110  ksi
Surface finish
Loading
surface  "ground"
load  "torsion"
Shape
Size (diameter)
round
d  2  in
Temperature
Reliability
T  72
Solution:
1.
R  0.999
See Mathcad file P0605a.
Using equation (6-5a), calculate the uncorrected endurance limit.
S'e 
return 0.5 S ut if S ut  200  ksi
S'e  55.0 ksi
100  ksi otherwise
2.
Calculate the endurance limit modification factors for a nonrotating round rod.
Load
Cload 
return 1 if load = "bending"
Cload  1
return 1 if load = "torsion"
return 0.7 if load = "axial"
Size
d equiv  d
 d equiv 

 in 
Surface
 0.097
Csize  0.869  
Csize  0.812
A 
A  1.34
return 1.34 if surface = "ground"
return 2.70 if surface = "machined"
return 2.70 if surface = "cold_rolled"
return 14.4 if surface = "hot_rolled"
return 39.9 if surface = "forged"
b 
return 0.085 if surface = "ground"
b  0.085
return 0.265 if surface = "machined"
return 0.265 if surface = "cold_rolled"
return 0.718 if surface = "hot_rolled"
return 0.995 if surface = "forged"
 S ut 

 ksi 
Temperature
b
Csurf  A  
Csurf  0.899
Ctemp 
Ctemp  1
return 1 if T  840
1  0.0032 ( T  840 ) otherwise
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
Creliab 
Reliability
6-5a-2
Creliab  0.753
return 1.000 if R = 0.50
return 0.897 if R = 0.90
return 0.814 if R = 0.99
return 0.753 if R = 0.999
return 0.702 if R = 0.9999
return 0.659 if R = 0.99999
3.
Calculate the modified endurance limit.
S e  Cload  Csize Csurf  Ctemp Creliab S'e
4.
S e  30.2 ksi
Using equation (6.9), calculate the fatigue strength at N = 10 3 cycles.
S m 
return 0.75 S ut if load = "axial"
S m  99.0 ksi
0.9 S ut otherwise
Sf = a N
b
5.
The equation for the S-N curve in the HCF region is given by equation (6.10a):
6.
Determine the constants a and b from equations (6.10c) and (6.10a). From Table 6-5, for N = 10 6 , z  3.000
b 
a 
1
z
 Sm 

 Se 
 log
Sm
10 
3
7.
b  0.1717
a  324.120 ksi
b
To draw the S-N graph over the range 10 3 <= N <= 10 8, define a piecewise continuous function.
S f ( N ) 
return a  N
b
if N  10
6
S e otherwise
8.
Plot the S-N curve over the range
3
5
N  10 10  10
8
100
Sf ( N )
ksi
10
3
1 10
4
1 10
5
6
1 10
1 10
7
1 10
8
1 10
N
FIGURE 6-5a
S-N Diagram for Problem 6-5a
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-6-1
PROBLEM 6-6
Statement:
For the trailer hitch from Problem 3-6 on p. 169 (also see Figures P6-2 and 1-5), find the infinite-lif
fatigue safety factors for all modes of failure assuming that the horizontal impact force of the traile
on the ball is fully reversed. Use steel with S ut = 600 MPa and S y = 450 MPa. Determine safety
factors for:
(a) The shank of the ball where it joins the ball bracket.
(b) Bearing failure in the ball bracket hole.
(c) Tearout failure in the ball bracket.
(d) Tensile failure in the 19-mm diameter attachment holes.
(e) Bending failure in the ball bracket as a cantilever.
Given:
a  40 mm
b  31 mm
Mtongue  100  kg Fpull  55.1 kN
S y  300  MPa
c  70 mm
d sh  26 mm
d  20 mm
t  19 mm
w  64 mm
R  32 mm
S ut  600  MPa
Assumptions: 1. The nuts are just snug-tight (no pre-load), which is the worst case.
2. All reactions will be concentrated loads rather than distributed loads or pressures.
Solution:
See Figures 6-6 and Mathcad file P0606.
W tongue
70 = c
1
F pull
1
40 = a
2
A
B
A
19 = t
B
F b1
31 = b
F a1x
C
F a1y
20 = d
F a2y
D
Fa2x
2
Fc2x
F b2
C
D
Fd2
F c2y
FIGURE 6-6A
Dimensions and Free Body Diagram for Problem 6-6
1.
The dynamic loading in this problem is fully reversed so the mean stresses are zero and the alternating stresses
are the same as those calculated in Problem 4-6. From Problem 4-6, the alternating components of the principal
stresses in the shank of the ball where it joins the ball bracket are:
σa1  1277 MPa
2.
σa2  0  MPa
Since 1 is the only nonzero principal stress, it is also the von Mises stress.
σa3  0  MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
σ'a  σa1
6-6-2
σ'a  1277 MPa
S'e  0.5 S ut
S'e  300 MPa
3.
Calculate the unmodified endurance limit.
4.
Calculate the endurance limit modification factors for a nonrotating round beam.
Load
Cload  1
Size
A95  0.010462 d sh
d equiv 
(bending load)
2
A95
A  4.51
Csurf
5.
 0.097
Csize  0.955
b  0.265
 Sut 
 A  

 MPa 
Temperature
Ctemp  1
Reliability
Creliab  0.753
2
d equiv  9.609 mm
0.0766
 d equiv 
Csize  1.189  

 mm 
Surface
A95  7.072 mm
(machined)
b
Csurf  0.828
(R = 99.9%)
Calculate the modified endurance limit.
S e  Cload  Csize Csurf  Ctemp Creliab S'e
S e  178.54 MPa
Na 
Se
Na  0.14
6.
Calculate the factor of safety for the ball shank.
7.
From Problem 4-6, the alternating components of the principal stresses at the bearing area in the ball bracket ho
are:
σa1  111.5  MPa
8.
σa2  0  MPa
σa3  0  MPa
Since 1 is the only nonzero principal stress, it is also the von Mises stress.
σ'a  σa1
9.
σ'a
σ'a  111.5 MPa
Calculate the endurance limit modification factors that are different from those in step 4.
Load
Cload  0.7
(axial load)
Size
Csize  1
(axial load)
10. Calculate the modified endurance limit.
S e  Cload  Csize Csurf  Ctemp Creliab S'e
11. Calculate the factor of safety for the bearing.
S e  130.91 MPa
Nb 
Se
σ'a
Nb  1.17
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-6-3
12. From Problem 4-6, the alternating components of the von Mises stress at the tearout shear area in the ball
bracket hole is:
σ'a  85.91  MPa
13. Calculate the endurance limit modification factors that are different from those in step 9.
Load
Cload  1
Size
A95  2  t ( 32 mm)   0.5 d sh
(simulated bending load)
2
d equiv 
2
A95  1111 mm
A95
2
d equiv  120.439 mm
0.0766
 d equiv 

 mm 
 0.097
Csize  1.189  
Csize  0.747
14. Calculate the modified endurance limit.
S e  Cload  Csize Csurf  Ctemp Creliab S'e
S e  139.71 MPa
Nc 
15. Calculate the factor of safety against tearout.
Se
Nc  1.6
σ'a
16. From Problem 4-6, the alternating components of the von Mises stress in the attachment bolts is:
σ'a  540.5  MPa
17. Calculate the endurance limit modification factors that are different from those in step 9.
Load
Cload  0.7
(axial load)
Size
Csize  1
(axial load)
18. Calculate the modified endurance limit.
S e  Cload  Csize Csurf  Ctemp Creliab S'e
19. Calculate the factor of safety against bolt tensile failure.
S e  130.91 MPa
Nd 
Se
Nd  0.24
σ'a
20. From Problem 4-6, the alternating components of the principal stresses in the cantilever beam are:
σa1  635.5  MPa
σa2  0  MPa
σa3  0  MPa
21. Since 1 is the only nonzero principal stress, it is also the von Mises stress.
σ'a  σa1
σ'a  635.5 MPa
22. Calculate the endurance limit modification factors that are different from those in step 17.
Load
Cload  1
Size
A95  0.05 w t
(bending load)
A95  60.8 mm
2
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
d equiv 
6-6-4
A95
d equiv  28.173 mm
0.0766
 d equiv 
Csize  1.189  

 mm 
 0.097
Csize  0.86
23. Calculate the modified endurance limit.
S e  Cload  Csize Csurf  Ctemp Creliab S'e
24. Calculate the factor of safety for the cantilever beam.
S e  160.85 MPa
Ne 
Se
σ'a
Ne  0.25
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-7-1
PROBLEM 6-7
Statement:
Design the wrist pin of Problem 3-7 for infinite life with a safety factor of 1.5 if the 2500-g
acceleration is fully reveresed and S ut = 130 ksi.
Given:
Force on wrist pin
Fwristpin  12.258 kN
Tensile strength
S ut  130  ksi
Design safety factor
Nd  1.5
od  0.375  in
Assumptions: Choose a suitable outside diameter, say
Solution:
Fwristpin  2756 lbf
See Figure 4-12 in the text and Mathcad file P0607.
Fwristpin
F 
F  1378 lbf
1.
The force at each shear plane is
2.
With only the direct shear acting on the plane, the Mohr diagram will be a circle with center at the origin and
radius equal to the shear stress. Thus, the principal normal stress is numerically equal to the shear stress,
which in this case is also the principal shear stress, so we have  = 1 = '.
3.
The shear stress at each shear plane is
τ=
2
F
A
=

4 F
2
= σ'

2
2
π  od  id   S
2
π od  id
Se
For fully reversed loading the factor of safety is, Nd =
5.
Calculate the unmodified endurance limit.
6.
Calculate the endurance limit modification factors for a nonrotating round pin (uniformly stressed).
Load
Cload  1
Size
A95 ( id) 
=
e
4.
σ'
S'e  0.5 S ut

2

7.
d equiv( id) 
4
A  1.34
Csurf
S'e  65 ksi
2
π od  id
 d equiv( id) 
Csize( id)  0.869  

 in 
Surface
4 F
Temperature
Ctemp  1
Reliability
Creliab  1.000
0.0766
 0.097
b  0.085
 S ut 
 A  

 ksi 
A95 ( id)
(ground)
b
Csurf  0.886
(R = 50%)
Calculate the modified endurance limit.
S e( id)  Cload  Csize( id)  Csurf  Ctemp Creliab S'e
8.
Solving for the inside diameter, guess id  0.2 in
Given
Nd =


2
2
π od  id  S e( id)
4 F
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
id  Find ( id)
9.
6-7-2
id  0.299 in
Round this down to the decimal equivalent of a common fraction (9/32),
id  0.281  in
10. The realized factor of safety is,
Nf 


2
2
π od  id  S e( id)
4 F
Nf  1.8
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-8-1
PROBLEM 6-8
Statement:
Given:
A paper machine processes rolls of paper having a density of 984 kg/m3. The paper roll is 1.50-m
OD x 0.22-m ID x 3.23 m long and is on a simply supported, hollow, steel shaft with S ut = 400 MPa
Find the shaft ID needed to obtain a dynamic safety factor of 2 for a 10-year life if the shaft OD is
22 cm and the roll turns at 50 rpm.
y
Paper roll:
ρ  984 
Density
Outside dia.
Inside dia.
Length
Shaft:
Strength
kg
w
3
m
OD  1500 mm
ID  220  mm
L  3230 mm
x
L
R
R
V
S ut  400  MPa
R
Outside dia.
Design safety factor
od  220  mm
Nfd  2
0
Design life
Shaft speed
Life  10 yr
ω  50 rpm
L/2
L
x
-R
M
2
wL /8
Assumptions: 1. The shaft is stiffer than the paper roll so the
weight of the roll on the shaft can be modelled as a
uniformly distributed load.
2. The bearings that support the shaft are close to
x
0
the ends of the paper roll and are thin with respect
L/2
L
to the length of the roll so we can consider the
distance between the shaft supports to be the
FIGURE 6-8
same as the length of the roll.
Load, Shear, and Moment Diagrams
3. There are no stress concentrations near the
for Problem 6-8
point of maximum moment on the shaft.
4. The paper mill operates 3 shifts/day, 365 days/year.
5. The shaft is machined and the material reliabilty is 99.9%.
See Figure 6-8 and Mathcad file P0608.
Solution:
1.
This is a case of fully reversed bending. The FBD for this loading case is shown in Appendix B, Figure B-2b,
with the dimension a equal to 0. That is, the distributed load starts at the left support and ends at the right
support.
2.
Calculate the number of stress cycles to see if we will design for finite or infinite life.
Nlife  Life ω
9
Nlife  1.652  10
cycles
This is well beyond 10 6 cycles, so we will design for infinite life.
3.
4.
Determine the weight of the paper roll and the magnitude of the distributed load on the shaft.

4
π
Roll volume
V 
Roll weight
W  V  ρ  g
Distributed
load on shaft
w 
2
2

 OD  ID  L
9
V  5.585  10  mm
3
W  53.895 kN
W
w  16.686
L
N
(a
mm
Figure D-2b shows that the maximum bending moment occurs at the center of the shaft and is
2
Mmax 
w L
8
7
Mmax  2.176  10  N  mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
Ma  Mmax
This is fully reversed bending so
5.
Mm  0  N  mm
and
The stress in the shaft at the point of maximum bending moment will depend upon the, as yet, unknown id. Tha
is,
I ( id) 
Area moment of inertia
π
64
σa( id) 
Alternating stress
6.
6-8-2

4

4
 od  id
(b)
Ma od
(c)
2  I ( id)
Calculate the modified endurance strength of the shaft.
S'e  0.5 S ut
Unmodified endurance limit
S'e  200  MPa
Modification factors:
Load
Cload  1
Size
Csize  1.189  


 mm 
od
 0.097
 Sut 
 4.51 

 MPa 
Surface
Csurf
Temperature
Ctemp  1
Reliability
Creliab  0.753
Csize  0.705
 0.265
Csurf  0.922
Modified endurance limit
S e  Cload  Csize Csurf  Ctemp Creliab S'e
7.
S e  97.8 MPa
(d)
Use the factor of safety equation as a design equation to solve for the unknown id. For fully reversed bending,
the factor of safety is
Nf =
Se
(e)
σa
Substituting equations b and c into e and solving for id,
1
4
 4 32 Nfd  Ma od 
id   od 

π S e


Rounding this, let the shaft ID be
id  190  mm
This gives a wall thickness of
t 
1
2
 ( od  id)
id  191.526  mm
t  15 mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-9-1
PROBLEM 6-9
Statement:
For the Vise Grip plier-wrench is drawn to scale in Figure P6-3, and for which the forces were
analyzed in Problem 3-9 and the stresses in Problem 4-9, find the safety factors for each pin for an
assumed clamping force of P = 4000 N in the position shown. The pins are 8-mm dia, S y = 400 MP
S ut = 520 MPa, and are all in double shear. Assume a desired finite life of 5E4 cycles.
Given:
Pin stresses as calculated in Problem 4-9:
Pin 1-2
τ12  74.6 MPa
S y  400  MPa
Yield strength
Pin 1-4
τ14  50.7 MPa
Tensile strength S ut  520  MPa
Pin 2-3
τ23  50.7 MPa
Pin diameter
d  8  mm
Pin 3-4
τ34  50.7 MPa
Desired life
Nlife  5  10
4
Assumptions: 1. Links 3 and 4 are in a toggle position, i.e., the pin that joins links 3 and 4 is in line with the pins
that join 1 with 4 and 2 with 3.
Solution:
1.
See Figure 6-9 and Mathcad file P0609.
The FBDs of the assembly and each individual link are shown in Figure 6-9. The dimensions, as scaled from
Figure P5-3 in the text, are shown on the link FBDs.
4
F
P
1
2
3
P
F
55.0 = b
50.0 = a
39.5 = c
F
F14
22.0 = d
129.2°
1

4
F34
F41
F21
P

28.0 = e


F43

F12
3
F23
F32
P
2.8 = g
21.2 = h
2
F
26.9 = f
FIGURE 6-9
Free Body Diagrams for Problem 6-9
2.
The pins are in pure shear, so the principal stresses are
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
3.
6-9-2
Pin joining 1 and 2
σ'12 
3  τ12
σ'12  129.211 MPa
All other pins
σ'14 
3  τ14
σ'14  87.815 MPa
This is a case of repeated fatigue loading. The alternating and mean von Mises stress components are:
Pin joining 1 and 2
σ'12a  0.5 σ'12
σ'12m  σ'12a
All other pins
σ'14a  0.5 σ'14
σ'14m  σ'14a
S'e  0.5 S ut
S'e  260 MPa
4.
Calculate the unmodified endurance limit.
5.
Calculate the endurance limit modification factors for a non rotating round pin (uniformly stressed).
Load
Cload  1
Size
A95 
π d
2
A95  50.265 mm
4
d equiv 
A95
d equiv  25.617 mm
0.0766
 d equiv 
Csize  1.189  

 mm 
A  4.51
Surface
 Sut 
 A  

 MPa 
Csurf
6.
Temperature
Ctemp  1
Reliability
Creliab  1.000
2
 0.097
Csize  0.868
b  0.265
(machined)
b
Csurf  0.86
(R = 50%)
Calculate the modified endurance limit.
S e  Cload  Csize Csurf  Ctemp Creliab S'e
S e  194.07 MPa
S m  0.9 S ut
S m  468 MPa
7.
Using equation (6.9), calculate the fatigue strength at N = 10 3 cycles.
8.
The equation for the S-N curve in the HCF region is given by equation (6.10a):
9.
Determine the constants a and b from equations (6.10c) and (6.10a). From Table 6-5, for N = 10 6 , z  3.000
b 
a 
1
z
 Sm 

 S'e 
 log
Sm
10 
3
b
Sf = a N
b
b  0.0851
a  842.4 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-9-3
4
10. Calculate the corrected fatigue strength at Nlife  5  10 cycles. S f  a  Nlife
b
S f  335.49 MPa
11. Assuming a Case 3 load line, use equation (6.18e) to determine the factor of safety.
Pin joining 1 and 2
Nf 
All other pins
Nf 
S f  S ut
σ'12a S ut  σ'12m S f
S f  S ut
σ'14a S ut  σ'14m S f
Nf  3.2
Nf  4.6
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-10-1
PROBLEM 6-10
Statement:
An overhung diving board is shown in Figure P6-4a. A 100-kg person is standing on the free end.
Assume cross-sectional dimensions of 305 mm x 32 mm. What is the fatigue safety factor for
infinite life if the material is brittle fiberglass with S f = 39 MPa @ N = 5E8 cycles and S ut = 130 MP
in the longitudinal direction?
2000 = L
Given:
Mass of person
Board dimensions
Load dimensions
Material properties
M  100  kg
w  305  mm
t  32 mm
b  700  mm
L  2000 mm
S ut  130  MPa
R1
P
R2
S f5E8  39 MPa
700 = b
FIGURE 6-10
Free Body Diagram for Problem 6-10
Assumptions: 1. The given fatigue strength is fully corrected.
2. There are no stress-concentrations near the point of maximum moment on the diving board.
Solution:
See Figure 6-10 and Mathcad file P0610.
1.
This is a case of repeated bending. The FBD for this loading case is shown in Appendix B, Figure B-3a, with th
dimension a equal to L. That is, the concentrated force F is at the end of the overhung beam.
2.
Determine the weight of the person on the end of the board.
W  M  g
Weight
3.
W  980.7  N
(a)
Figure B-3a in Appendix B shows that the maximum bending moment occurs at the right-hand support and is
6
Mmax  W  ( L  b )
Mmax  1.275  10  N  mm
This is repeated bending so
Ma 
4.
Mmax
The stress in the board at the point of maximum bending moment is
Area moment of inertia
5.
Mm  Ma
and
2
I 
w t
Alternating stress
σa 
Mean stress
σm 
3
12
Ma t
2 I
Mm t
2 I
5
I  8.329  10  mm
4
(b)
σa  12.2 MPa
(c)
σm  12.2 MPa
(d)
For repeated (fluctuating) bending, the factor of safety for Case 3 loading is
Nf 
S f5E8 S ut
σa S ut  σm S f5E8
Nf  2.4
(e)
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-11-1
PROBLEM 6-11
Statement:
Repeat Problem 6-10 assuming the 100-kg person in Problem 6-10 jumps up 25 cm and lands back
on the board. Assume the board weighs 29 kg and deflects 13.1 cm statically when the person
stands on it. What is the fatigue safety factor for finite life if the material is brittle fiberglass with S
= 39 MPa @ N = 5E8 cycles and S ut = 130 MPa in the longitudinal direction?
Given:
Maximum principal stresses due to
bending at R2 from Problem 4-11
σ1max  76.3 MPa
2000 = L
R1
P
σ2max  0  MPa
σ3max  0  MPa
Ultimate strength
S ut  130  MPa
Fatigue strength
S f  39 MPa
Fatigue life
Ncycle  5  10
R2
8
700 = b
FIGURE 6-11
Solution:
1.
See Figure 6-11 and Mathcad file P0611.
Free Body Diagram for Problem 6-11
The dynamic loading in this case is repeated, i.e., the stresses go from zero to the maximum values given above.
Thus, the minimum and maximum von Mises stresses are:
σ'max 
2
σ1max  σ1max σ3max  σ3max
2
σ'max  76.3 MPa
2.
The alternating and mean components of the von Mises stress are:
σ'a 
σ'm 
3.
σ'min  0  MPa
σ'max  σ'min
2
σ'max  σ'min
2
σ'a  38.1 MPa
σ'm  38.1 MPa
Assuming a Case 3 load line, use equation (6.18e) to calculate the factor of safety.
Nf 
S f  S ut
σ'a S ut  σ'm S f
Nf  0.79
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-12-1
PROBLEM 6-12
Statement:
Repeat Problem 6-10 using the cantilevered diving board design in Figure P6-4b.
Given:
Mass of person
Board dimensions
2000
Load dimensions
Material properties
M  100  kg
w  305  mm
t  32 mm
L  1300 mm
S ut  130  MPa
1300 = L
P
M1
S f5E8  39 MPa
R1
Assumptions: 1. The given fatigue strength is fully corrected.
700
2. There are no stress-concentrations near the
point of maximum moment on the diving board. FIGURE 6-12
Free Body Diagram for Problem 6-12
Solution:
See Figure 6-12 and Mathcad file P0612.
1.
This is a case of repeated bending. The FBD for this loading case is shown in Appendix B, Figure B-1a, with
the dimension a equal to L. That is, the concentrated force F is at the end of the cantilever beam.
2.
Determine the weight of the person on the end of the board.
W  M  g
Weight
3.
W  980.7  N
(a)
Figure B-1a in Appendix B shows that the maximum bending moment occurs at the support and is
6
Mmax  W  L
Mmax  1.275  10  N  mm
This is repeated bending so
Ma 
4.
Mmax
The stress in the board at the point of maximum bending moment is
Area moment of inertia
5.
Mm  Ma
and
2
I 
w t
Alternating stress
σa 
Mean stress
σm 
3
12
Ma t
2 I
Mm t
2 I
5
I  8.329  10  mm
4
(b)
σa  12.2 MPa
(c)
σm  12.2 MPa
(d)
For repeated (fluctuating) bending, the factor of safety for Case 3 loading is
Nf 
S f5E8 S ut
σa S ut  σm S f5E8
Nf  2.4
(e)
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-13-1
PROBLEM 6-13
Statement:
Repeat Problem 6-11 using the cantilevered diving board design in Figure P6-4b. Assume the
board weighs 19 kg and deflects 8.5 cm statically when the person stands on it.
Given:
Maximum principal stresses due to
bending at support from Problem 4-13
σ1max  87.1 MPa
2000
1300 = L
P
σ2max  0  MPa
σ3max  0  MPa
Ultimate strength
S ut  130  MPa
Fatigue strength
S f  39 MPa
Fatigue life
Ncycle  5  10
M1
8
R1
700
FIGURE 6-13
Solution:
1.
See Figure 6-13 and Mathcad file P0613.
Free Body Diagram for Problem 6-13
The dynamic loading in this case is repeated, i.e., the stresses go from zero to the maximum values given above.
Thus, the minimum and maximum von Mises stresses are:
σ'max 
2
σ1max  σ1max σ3max  σ3max
2
σ'max  87.1 MPa
2.
The alternating and mean components of the von Mises stress are:
σ'a 
σ'm 
3.
σ'min  0  MPa
σ'max  σ'min
2
σ'max  σ'min
2
σ'a  43.5 MPa
σ'm  43.5 MPa
Assuming a Case 3 load line, use equation (6.18e) to calculate the factor of safety.
Nf 
S f  S ut
σ'a S ut  σ'm S f
Nf  0.69
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-14-1
PROBLEM 6-14
Statement:
Figure P6-5 shows a child's toy called a pogo stick. The child stands on the pads, applying half
her weight on each side. She jumps off the ground, holding the pads up against her feet, and
bounces along with the spring cushioning the impact and storing energy to help each rebound.
Design the aluminum cantilever beam sections on which she stands to survive jumping 2 in off
the ground with a dynamic safety factor of 2 for a finite life of 5E4 cycles. Use 2000 series
aluminum. Define and size the beam shape.
Given:
Heat treated 2024 aluminum:
Tensile strength
S ut  64 ksi
Design safety factor
Nfd  2
Design life
Nlife  5  10
4
Assumptions: The beam will have a rectangular crosssection with the load applied at a distance
of 5 in from the central support.
L  5  in
Solution:
1.
See Figure 6-14 and Mathcad file P0614.
Fi /2
From Problem 3-14, the total dynamic force on both
foot supports is
Fimax  224  lbf
Fi /2
Fimin  0  lbf
Therefore, the load on each support is
Pmax 
Pmin 
Fimax
Pmax  112  lbf
2
Fimin
Pmin  0  lbf
2
2.
To give adequate support to the childs foot, let the
width of the support beam be w  1.5 in
3.
From Figure B-1(a) in Appendix B, the maximum
bending moment at x = 0 is
Mmax  Pmax L
4.
P
FIGURE 5-14
Free Body Diagram for Problem 5-14
Mmax  560  in lbf
Mmin  0  in lbf
Calculate the alternating and mean components of the bending moment.
Ma 
Mm 
Mmax  Mmin
Ma  280  in lbf
2
Mmax  Mmin
Mm  280  in lbf
2
S'e  19 ksi @ 5E8 cycles
5.
Determine the unmodified endurance limit.
6.
Calculate the endurance limit modification factors for a nonrotating rectangular beam.
Load
Cload  1
Size
A95 ( t)  0.05 w t
d equiv( t) 
A95 ( t)
0.0766
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-14-2
 d equiv( t) 
Csize( t)  0.869  

 in 
A  2.7
Surface
Csurf
7.
 0.097
b  0.265
 S ut 
 A  

 ksi 
Temperature
Ctemp  1
Reliability
Creliab  1.000
(machined)
b
Csurf  0.897
(R = 50%)
Calculate the modified endurance limit.
S e( t)  Cload  Csize( t)  Csurf  Ctemp Creliab S'e
S m  0.9 S ut
8.
Using equation (6.9), calculate the fatigue strength at N = 10 3 cycles.
9.
The equation for the S-N curve in the HCF region is given by equation (6.10a):
S m  57.6 ksi
Sf = a N
b
10. Determine the constants a and b from equations (6.10c) and (6.10a). From Table 6-5, for N = 5E8 , z  5.699
1
b ( t) 
z
 Sm 

 Se( t) 
 log
a ( t) 
Sm
103
b( t )
4
11. Determine the corrected fatigue strength at Nlife  5  10 cycles.
S f ( t)  a ( t)  Nlife
b( t )
12. We can now determine the minimum required section depth, t. Using the distortion-energy failure theory
with the modified Goodman diagram, the bending stress will also be the only nonzero principal stress,
which will also be the von Mises stress. Assuming a Case 3 load line, use equation (6.18e) to determine the
factor of safety. Guess t  10 mm.
Bending stress
σ=
M c
I
Given
Nfd =
t 12
6 M
= M 
=
2
3
2
w t
w t
w t
6
2

S f ( t)  S ut
Ma S ut  Mm S f ( t)
t  Find ( t)
t  0.304  in
Round this up to the next higher decimal equivalent of a common fraction,
t  0.375  in
Using this value of t, the values of the functions of t are:
Csize( t)  0.912
S e( t)  15.545 ksi
S f ( t)  38.981 ksi
The realized safety factor is
Nf 
w t
6
2

S f ( t)  S ut
Ma S ut  Mm S f ( t)
Nf  3.0
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-15a-1
PROBLEM 6-15a
Statement:
For a notched part having a notch dimension r, geometric stress concentration factor Kt, and
material strength S ut as shown in row a of Table P6-4, find the Neuber factor a, the material's notch
sensitivity q, and the fatigue stress-concentration factor Kf.
Given:
Ultimate tensile strength
S ut  100  ksi
Geometric stress-concentration factor
Kt  3.3
Notch radius
r  0.25 in
Material is steel
Loading  "Bending"
Solution:
See Mathcad file P0615a.
1
2
1.
From Table 6-6, the Neuber constant for S ut  100 ksi is
a  0.062  in
2.
Using equation 6-13, the notch sensitivity is
q 
1
1
3.
a  0.062 in
2
q  0.890
a
r
The fatigue stress-concentration factor, from equation 6.11b, is
Kf  1  q   Kt  1 
Kf  3.05
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P0615a.xmcd
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-16-1
PROBLEM 6-16
Statement:
A track to guide bowling balls is designed with two round rods as shown in Figure P6-6. The
rods are not parallel to one another but have a small angle between them. The balls roll on the
rods until they fall between them and drop onto another track. The angle between the rods is
varied to cause the ball to drop at different locations. Find the infinite-life safety factor for the
1-in dia SAE 1010 cold-rolled steel rods.
(a) Assume rods are simply supported at each end.
(b) Assume rods are fixed at each end.
Given:
Tensile strength
S ut  53 ksi
Rod diameter
d  1.00 in
Solution:
Fball
a
See Figure 6-16 and Mathcad file P0616.
R1
1.
The maximum bending stress will occur at the outer
fibers of the rod at the section where the maximum
bending moment occurs which, in this case, is at x = a.
The only stress present on the top or bottom surface
of the rod is the bending stress x. Therefore, on the
bottom surface where the stress is tensile, sx is the
principal stress 1 . Thus, from Problem 4-16, for a
simply supported rod,
Maximum principal stress
2.
R2
L
FIGURE 6-16A
Free Body Diagram for Problem 6-16(a), taken
on a plane through the rod axis and ball center
σ1  748  psi
The dynamic loading is repeated from 0 to 1 for each ball that rolls down the track. The alternating and mean
components of the von Mises stress are:
Alternating von Mises stress
σ'a  0.5 σ1
σ'a  374 psi
Mean von Mises stress
σ'm  0.5 σ1
σ'm  374 psi
S'e  0.5 S ut
S'e  26.5 ksi
3.
Calculate the unmodified endurance limit.
4.
Calculate the endurance limit modification factors for a nonrotating round beam.
Load
Cload  1
Size
A95  0.010462 d
2
d equiv 
 d equiv 
Csize  0.869  

 in 
Surface
A  2.70
Csurf
 S ut 
 A  

 ksi 
Temperature
Ctemp  1
Reliability
Creliab  0.659
A95
0.0766
 0.097
b  0.265
Csize  0.957
(machined)
b
Csurf  0.943
(R = 99.999%)
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5.
6-16-2
Calculate the modified endurance limit.
S e  Cload  Csize Csurf  Ctemp Creliab S'e
6.
Assuming a Case 3 load line, use equation (6.18e) to determine the factor of safety.
Case a
7.
Nfa 
S e S ut
9.
Nfa  32
σ'a S ut  σ'm S e
For the built-in case, the maximum bending stress
will occur at the outer fibers of the rod at the
section where the maximum bending moment
occurs which, in this case, is at x = L. The only
stress present on the top or bottom surface of the
rod is the bending stress x. Therefore, on the
bottom surface where the stress is tensile, sx is the
principal stress 1 . Thus, from Problem 4-16, for a
simply supported rod,
Fball
a
M1
R1
L
R 2 M2
FIGURE 6-16B
Free Body Diagram for Problem 6-16(b), taken on a
plane through the rod axis and ball center
σ1  577  psi
Maximum principal stress
8.
S e  15.759 ksi
The dynamic loading is repeated from 0 to 1 for each ball that rolls down the track. The alternating and mean
components of the von Mises stress are:
Alternating von Mises stress
σ'a  0.5 σ1
σ'a  288.5 psi
Mean von Mises stress
σ'm  0.5 σ1
σ'm  288.5 psi
Assuming a Case 3 load line, use equation (6.18e) to determine the factor of safety.
Case b
Nfb 
S e S ut
σ'a S ut  σ'm S e
Nfb  42
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-17-1
PROBLEM 6-17
Statement:
A pair of ice tongs is shown in Figure P6-7. The ice weighs 50 lb and is 10 in wide across the
tongs. The distance between the handles is 4 in, and the mean radius r of the tong is 6 in. The
rectangular cross-sectional dimensions are 0.75 x 0.312 in. Find the safety factor for the tongs for
5E5 cycles if their S ut = 50 ksi.
F
Given:
Tensile strength
S ut  50 ksi
Cross-section:
Width
Depth
w  0.312  in
h  0.75 in
Life
Nf  5  10
C
FC
O
5
11.0 = ax
3.5 = cy
FO
2.0 = cx
A
12.0 = by
Assumptions: The tongs are forged. Use 99.99% reliability.
Operating temperature is between 32F and 70F.
5.0 = bx
FB
Solution:
See Problem 4-17, Figure 6-17, and
Mathcad file P0617.
B
1. The maximum bending stress in the tong was found
in Problem 4-17 at point A.
Vertical direction
W/2
FIGURE 6-17
Free Body Diagram for Problem 6-17
σi  8.58 ksi
All other components are zero.
2.
There are no other stress components present so
σ1max  σi
3.
σ1max  8.58 ksi
σ3max  0  ksi
The dynamic loading in this case is repeated, thus
σ1min  0  ksi
4.
σ2max  0  ksi
σ2min  0  ksi
σ3min  0  ksi
Even though this is a brittle material, for HCF analysis, determine the von Mises effective stresses. Since there i
only one nonzero stress,
σ'max  σ1max
σ'max  8.58 ksi
σ'min  σ1min
σ'min  0 ksi
σ'a 
σ'm 
σ'max  σ'min
σ'a  4.29 ksi
2
σ'max  σ'min
σ'm  4.29 ksi
2
S'e  0.5 S ut
S'e  25 ksi
5.
Calculate the unmodified endurance limit.
6.
Calculate the endurance limit modification factors for a nonrotating rectangular beam.
Load
Cload  1
Size
A95  0.05 w h
d equiv 
A95
0.0766
A95  7.548 mm
2
d equiv  9.927 mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-17-2
 d equiv 
Csize  1.189  

 mm 
A  39.9
Surface
 S ut 
 A  

 ksi 
Csurf
7.
Temperature
Ctemp  1
Reliability
Creliab  0.702
 0.097
Csize  0.952
b  0.995
(forged)
b
Csurf  0.814
(R = 99.99%)
Calculate the modified endurance limit.
S e  Cload  Csize Csurf  Ctemp Creliab S'e
S e  13.59 ksi
= 10 3
S m  0.9 S ut
8.
Using equation (6.9), calculate the fatigue strength at N
cycles.
9.
The equation for the S-N curve in the HCF region is given by equation (6.10a):
Sf = a N
S m  45 ksi
b
10. Determine the constants a and b from equations (6.10c) and (6.10a). From Table 6-5, for N = 10 6 , z  3.000
b 
a 
1
z
 Sm 

 Se 
 log
Sm
10 
3
b
11. Using equation (6.10a), determine the fatigue strength.
b  0.1733
a  148.991 ksi
S f5E5  a  Nf
b
S f5E5  15.326 ksi
12. Assuming a Case 3 load line, use equation (6.18e) to calculate the factor of safety.
Nf5E5 
S f5E5 S ut
σ'a S ut  σ'm S f5E5
Nf5E5  2.7
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-18-1
PROBLEM 6-18
Statement:
A pair of ice tongs is shown in Figure P6-7. The ice weighs 50 lb and is 10 in wide across the
tongs. The distance between the handles is 4 in, and the mean radius r of the tong is 6 in. The
rectangular cross-sectional dimensions are 0.75 x 0.312 in. Find the safety factor for the tongs
for 5E5 cycles if they are made of Class 40 gray cast iron.
Given:
Tensile strength
S ut  42 ksi
Cross-section:
Width
Depth
w  0.312  in
h  0.75 in
Nf  5  10
Life
F
C
FC
3.5 = cy
O
5
FO
11.0 = ax
Assumptions: The tongs are as-cast. Use 99.99%
reliability. Operating temperature is
between 32F and 70F. Set Csurf to 1 for a
cast finish, which does not need a
surface factor.
See Problem 4-18, Figure 6-18, and
Solution:
Mathcad file P0618.
2.0 = cx
A
12.0 = by
5.0 = bx
FB
B
W/2
1.
The maximum bending stress in the tong was found in
Problem 4-17 at point A.
σi  8.58 ksi
Vertical direction
FIGURE 6-18
Free Body Diagram for Problem 6-18
All other components are zero.
2.
There are no other stress components present so
σ1max  σi
3.
σ1max  8.58 ksi
σ3max  0  ksi
The dynamic loading in this case is repeated, thus
σ1min  0  ksi
4.
σ2max  0  ksi
σ2min  0  ksi
σ3min  0  ksi
Even though this is a brittle material, for HCF analysis, determine the von Mises effective stresses. Since
there is only one nonzero stress,
σ'max  σ1max
σ'max  8.58 ksi
σ'min  σ1min
σ'min  0 ksi
σ'a 
σ'm 
σ'max  σ'min
σ'a  4.29 ksi
2
σ'max  σ'min
σ'm  4.29 ksi
2
S'e  0.4 S ut
S'e  16.8 ksi
5.
Calculate the unmodified endurance limit.
6.
Calculate the endurance limit modification factors for a nonrotating rectangular beam.
Load
Cload  1
Size
A95  0.05 w h
A95  7.548 mm
2
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
d equiv 
A95
d equiv  9.927 mm
0.0766
 d equiv 
Csize  1.189  

 mm 
7.
6-18-2
Surface
Csurf  1
Temperature
Ctemp  1
Reliability
Creliab  0.702
 0.097
Csize  0.952
(R = 99.99%)
Calculate the modified endurance limit.
S e  Cload  Csize Csurf  Ctemp Creliab S'e
S e  11.22 ksi
S m  0.9 S ut
8.
Using equation (6.9), calculate the fatigue strength at N = 10 3 cycles.
9.
The equation for the S-N curve in the HCF region is given by equation (6.10a):
Sf = a N
S m  37.8 ksi
b
10. Determine the constants a and b from equations (6.10c) and (6.10a). From Table 6-5, for N = 10 6 , z  3.000
b 
a 
1
z
 Sm 

 Se 
 log
Sm
10 
3
b
11. Using equation (6.10a), determine the fatigue strength.
b  0.1758
a  127.305 ksi
S f5E5  a  Nf
b
S f5E5  12.678 ksi
12. Assuming a Case 3 load line, use equation (6.18e) to calculate the factor of safety.
Nf5E5 
S f5E5 S ut
σ'a S ut  σ'm S f5E5
Nf5E5  2.3
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-19-1
PROBLEM 6-19
Statement:
Determine the size of the clevis pin shown in Figure P6-8 needed to withstand an applied repeated
force of 0 to 130000 lb for infinite life. Also determine the required outside radius of the clevis
end to not fail in either tearout or bearing if the clevis flanges are each 2.5 in thick. Use a safety
factor of 3. Assume S ut = 140 ksi for the pin and S ut = 80 ksi for the clevis.
Given:
Minimum force
Pmin  0  lbf
Maximum force
Pmax  130  kip
Pin
S utp  140  ksi
Flange thickness
t  2.5 in
Clevis
S utc  80 ksi
Material strength:
Nf  3
Factor of safety against fatigue failure
Assumptions: The parts are machined. Use 90% reliability and room temperature.
Solution:
1.
See Figure 6-19 and Mathcad file P0619.
Calculate the alternating and mean components of the forces on the clevis and link.
Pa 
Pm 
Pmax  Pmin
Pa  65 kip
2
Pmax  Pmin
Pm  65 kip
2
Stress in Pin
2.
The pin is in double shear and there is no stress-concentration. The alternating and mean loads at one section
on the pin are
τa =
1 Pa

2 Apin
τm =
Apin ( d ) 
1 Pm

2 Apin
π d
2
(2)
3.
The cross-section area of the pin is
4.
The alternating and mean shear stresses and von Mises stresses are
τa( d ) 
σ'a( d ) 
Pa
4
Pm
τm( d ) 
2  Apin ( d )
3  τa( d )
(1)
(3)
2  Apin ( d )
σ'm( d ) 
3  τm( d )
(4)
Pin Strength
S'ep  0.5 S utp
S'ep  70 ksi
5.
Calculate the unmodified endurance limit.
6.
Calculate the endurance limit modification factors for a nonrotating round pin in bending.
Load
Cload  1
Size
A95 ( d )  0.010462 d
2
 dequiv( d) 

 in 
d equiv( d ) 
A95 ( d )
0.0766
 0.097
Csize( d )  0.869  
Surface
A  2.70
b  0.265
(machined)
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
 S utp 
 A  

 ksi 
Csurf
7.
Temperature
Ctemp  1
Reliability
Creliab  0.897
6-19-2
b
Csurf  0.729
(R = 90%)
Calculate the modified endurance limit.
S ep( d )  Cload  Csize( d )  Csurf  Ctemp Creliab S'ep
(5)
Design Equation
8.
Using the modified-Goodman failure criterion and a case 3 load line, the factor of safety is given by equation
6-18e as
Nf =
9.
S e S ut
(6)
σ'a S ut  σ'm S e
Substituting equations 4 and 5 into 6 and solving for d yields
Given
Nf =
Guess
d  2.0 in
S ep( d )  S utp
σ'a( d )  S utp  σ'm( d )  S ep( d )
d  Find ( d )
d  2.632 in
d  2.750  in
Rounding to the next higher eighth of an inch, let
With this value of d, we have
σ'a( d )  9.5 ksi
σ'm( d )  9.5 ksi
S ep( d )  39.71 ksi
and the realized factor of safety against fatigue failure in the pin is
Nf 
S ep( d )  S utp
Nf  3.3
σ'a( d )  S utp  σ'm( d )  S ep( d )
Tearout length
Clevis Tearout (See Figure 6-19)
10. Let the outside radius of the clevis be R. Then the tearout
area is
2
Atear ( R)  2  t  R  ( 0.5 d )
2
11. The alternating and mean shear stresses and von Mises
stresses are
τa( R) 
Pa
2  Atear ( R)
τm( R) 
Pm
2  Atear ( R)
d
(7)
R
FIGURE 6-19
Tearout Diagram for Problem 6-19
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
σ'a( R) 
3  τa( R)
σ'm( R) 
6-19-3
3  τm( R)
(8)
Clevis Strength
12. Calculate the unmodified endurance limit.
S'ec  0.5 S utc
S'ec  40 ksi
13. Calculate the endurance limit modification factors for a nonrotating rectangular shear area (uniformly stressed).
Load
Cload  1
Size
A95 ( R)  Atear ( R)
d equiv( R) 
 d equiv( R) 
Csize( R)  0.869  

 in 
A  2.70
Surface
Csurf
Temperature
Ctemp  1
Reliability
Creliab  0.897
0.0766
 0.097
b  0.265
 S utc 
 A  

 ksi 
A95 ( R)
(machined)
b
Csurf  0.845
(R = 90%)
14. Calculate the modified endurance limit.
S ec( R)  Cload  Csize( R)  Csurf  Ctemp Creliab S'ec
(9)
Design Equation
15. Using the modified-Goodman failure criterion and a case 3 load line, the factor of safety is given by equation
6-18e as
S e S ut
(10)
Nf =
σ'a S ut  σ'm S e
16. Substituting equations 8 and 9 into 10 and solving for R yields
Given
Nf =
Guess
R  2  in
S ec( R)  S utc
σ'a( R)  S utc  σ'm( R)  S ec( R)
R  Find ( R)
R  2.624 in
R  2.625  in
Bearing Stress
The maximum bearing stress in the hole in each flange is
σmaxbear 
 Pa  Pm
2 d t
σmaxbear  9.5 ksi
This is small compared to the ultimate strength of the clevis.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-20-1
PROBLEM 6-20
Statement:
A ±100 N-m torque is applied to a 1-m-long, solid round steel shaft. Design it to limit its angular
deflection to 2 deg and select a steel alloy to have a fatigue safety factor of 2 for infinite life.
Given:
Applied torque
Ta  100  N  m
Shaft length
Max deflection
L  1000 mm
θmax  2  deg
Design safety factor
Nfd  2
Modulus of rigidity
G  80.8 GPa
Tm  0  N  m
Assumptions: There are no stress-concentrations anywhere on the shaft. The shaft is machined, reliability is
99.9%, and the it is at room temperature.
Solution:
1.
See Mathcad file P0620.
This is a case of fully reversed torsion. We will use the von Mises effective stress so the load factor will be 1.
Tmax  Ta  Tm
The maximum torque is
2.
Tmax  100 N  m
The diameter of the shaft can be found from equations 4.24 and 4.25 with  = max.
θmax =
Tmax L
J G
=
32 Tmax L
4
π d  G
1
3.
Solving for d,
 32 Tmax L 
d  

 π θmax G 
Rounding, let
d  24.5 mm
4
d  24.514 mm
Now, we can solve for the stress in the shaft.
Polar moment
of inertia
J 
Torsional stress
τa 
π
32
d
4
Ta d
2 J
4
J  3.537  10 mm
4
τa  34.632 MPa
The corresponding von Mises normal stress is
von Mises stress
4.
3  τa
σ'a  59.984 MPa
Using the factor of safety equation for reversed loading, calculate the required endurance limit
Nf =
5.
σ'a 
Se
σ'a
S e  Nfd  σ'a
S e  119.967 MPa
This endurance limit is a function of the unknown ultimate tensile strength. Use the endurance limit modificati
equation to determine the required S ut.
S e = Cload  Csize Csurf  Ctemp Creliab S'e
6.
Calculate the endurance limit modification factors for a solid, round steel shaft.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
Load
Cload  1
Size
Csize  1.189  
Surface
A  4.51


 mm 
Csurf
7.
6-20-2
d
 0.097
Csize  0.872
b  0.265
 S ut 
= A 

 MPa 
Temperature
Ctemp  1
Reliability
Creliab  0.753
Uncorrected
endurance strength
S'e = 0.5 S ut
(machined)
b
(R = 99.9%)
Substituting these into the equation above and solving for S ut,
1
Se


S ut  

 0.5 A  Csize Creliab MPa 
b 1
 MPa
S ut  395 MPa
Based on this requirement, choose AISI 1020 cold-rolled steel that will be machined to size.
8.
Check the actual factor of safety based on the material chosen. For this material, S ut  469  MPa
 Sut 
 A  

 MPa 
b
Surface factor
Csurf
Uncorrected
endurance strength
S'e  0.5 S ut
Corrected
endurance strength
S e  Cload  Csize Csurf  Ctemp Creliab S'e
Csurf  0.884
S'e  234.5 MPa
S e  136.0 MPa
Factor of safety
Nf 
Se
σ'a
Nf  2.3
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-21-1
PROBLEM 6-21
Statement:
Figure P6-9 shows an automobile wheel with two common styles of lug wrench being used to
tighten the wheel nuts, a single-ended wrench in (a), and a double-ended wrench in (b). The
distance between points A and B is 1 ft in both cases and the handle diameter is 0.625 in. How
many cycles of tightening can be expected before a fatigue failure if the average tightening torque
is 100 ft-lb and the material S ut = 60 ksi?
Given:
Distance between A and B
d AB  1  ft
Minimum torque Tmin  0  ft  lbf
Wrench diameter
d  0.625  in
Maximum torque Tmax  100  ft  lbf
Tensile strength
S ut  60 ksi
Assumptions: 1. The forces exerted by the user's hands lie in a plane through the wrench that is also parallel to
the plane of the wheel.
2. The applied torque is perpendicular to the plane of the forces.
3. By virtue of 1 and 2 above, this is a planar problem that can be described in a 2D FBD.
4. The surface is as-forged, the reliability is 50%, and the wrench will not be used in extremely ho
or cold environments.
Solution:
1.
See Figure 6-21 and Mathcad file P0621.
12" = dAB
From examination of the FBDs, we see that, in
both cases, the arms are in bending and the
stub that holds the socket wrench is in pure
torsion. The maximum bending stress in the
arm will occur near the point where the arm
transitions to the stub. The stress state at
this transition is very complicated, but we can
find the nominal bending stress there by
treating the arm as a cantilever beam, fixed at
the transition point. For both cases the
torque in the stub is the same.
F
T
F
(a) Single-ended Wrench
12" = dAB
Case (a)
2.
F
6"
The bending moment at the transition is
M = F  d AB = T
Ma 
T
Tmax  Tmin
F
2
(b) Double-ended Wrench
Ma  600 in lbf
FIGURE 6-21
Mm  Ma
3.
Free Body Diagrams for Problem 6-21
The alternating and mean components of the bending stress at this point are found from
Moment of inertia
I 
π d
Dist to extreme fibre
64
c  0.5 d
Alternating stress
σxa 
Mean stress
σxm 
4
I  in
4
c  0.313 in
Ma c
I
Mm c
I
σxa  25.033 ksi
σxm  25.033 ksi
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4.
There are no other stress components present at this point, so x is the maximum principle stress here and
σ2  0  psi
σ1 = σx
5.
σ3  0  psi
Since there is only one nonzero principal stress, the von Mises stress is
σ'a  σxa
6.
6-21-2
σ' = σ1 = σx
σ'm  σxm
Assuming a Case 3 load line, use equation (6.18e) to solve for the fatigue strength at which the wrench will fail
(safety factor of 1).
Nf =
S f  S ut
σ'a S ut  σ'm S f
=1
S f 
σ'a S ut
S f  42.954 ksi
S ut  σ'm
7.
Using equation (6-5a), calculate the uncorrected endurance limit. S'e  0.5 S ut
8.
Calculate the endurance limit modification factors for a nonrotating round beam.
Load
Cload  1
Size
A95  0.010462 d
2
 d equiv 
Csize  0.869  

 in 
A  39.9
Surface
Csurf
9.
and
d equiv 
Temperature
Ctemp  1
Reliability
Creliab  1.000
A95
0.0766
 0.097
Csize  1.002
b  0.995
 S ut 
 A  

 ksi 
S'e  30 ksi
(as forged)
b
Csurf  0.679
(R = 50%)
Calculate the modified endurance limit.
S e  Cload  Csize Csurf  Ctemp Creliab S'e
10. Using equation (6.9), calculate the fatigue strength at N = 10 3 cycles.
S e  20.398 ksi
S m  0.9 S ut
11. The equation for the S-N curve in the HCF region is given by equation (6.10a):
Sf = a N
S m  54 ksi
b
12. Determine the constants a and b from equations (6.10c) and (6.10a). From Table 6-5, for N = 10 6 , z  3.000
b 
a 
1
z
 Sm 

 Se 
 log
Sm
10 
3
b
b  0.1409
a  142.955 ksi
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-21-3
1
13. Calculate the number of cycles to failure using equation (6.10a)
 Sf 
Na   
a
b
3
Na  5.1  10
Case (b)
15. The tensile stress at this point is found from
F  d AB
M=
14. The bending moment at the transition is
σx =
M c
I
2
=
T
2
Tc
=
2 I
16. The bending stress in the handle for case (b) is one half that of case (a). However, the torque in the stub is the
same in both cases. The shear stress at any point on the outside surface of the stub is found from
Polar moment of inertia
J  2  I
Maximum shear stress
τxymax 
τxymin 
Minimum shear stress
τa 
Alternating shear stress
τm 
Mean shear stress
4
J  0.0150 in
Tmax c
τxymax  25.03 ksi
J
Tmin c
τxymin  0 ksi
J
τxymax  τxymin
τa  12.52 ksi
2
τxymax  τxymin
τm  12.52 ksi
2
17. There are no other stress components present along the outside surface of the stub, so
σ1a  τa
and
σ'a 
σ1a  12.5 ksi
2
σ2a  0  psi
2
σ1a  σ1a σ3a  σ3a
σ3a  σ1a
σ'a  21.7 ksi
σ'm  σ'a
σ'm  21.7 ksi
18. Assuming a Case 3 load line, use equation (6.18e) to solve for the fatigue strength at which the wrench will fail
(safety factor of 1).
Nf =
S f  S ut
σ'a S ut  σ'm S f
=1
S f 
σ'a S ut
S f  33.944 ksi
S ut  σ'm
1
19. Calculate the number of cycles to failure using equation (6.10a)
Nb 
 Sf 
 
a
b
4
Nb  2.7  10
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-22-1
PROBLEM 6-22
Statement:
A roller-blade skate is shown in Figure P6-10. The polyurethane wheels are 72 mm dia and spaced
on 104-mm centers. The skate-boot-foot combination weighs 2 kg. The effective "spring rate" of
the person-skate subsystem is 6000 N/m. The axles are 10-mm-dia steel pins in double shear with
S ut = 550 MPa. Find the fatigue safety factor for the pins when a 100-kg person lands a 0.5-m jump
on one foot assuming infinite life.
(a) Assume all 4 wheels land simultaneously.
(b) Assume that one wheel absorbs all the landing force.
Given:
Axle pin diameter
d  10 mm
Tensile strength S ut  550  MPa
Assumptions: Pins are machined and reliability is 99.999%.
Solution:
See Figure P6-10 and Mathcad file P0622.
τa  5.71 MPa
τb  22.9 MPa
1.
From Problem 4-22, we have the stresses for cases (a) and (b):
2.
The dynamic loading in this case is repeated so the stresses given in step 1 are the maximum and the minimum
stresses are zero. Determine the minimum, maximum, alternating, and mean von Mises stresses.
Part (a)
σ'maxa 
σ'aa 
σ'ma 
Part (b)
σ'mb 
σ'maxa  9.89 MPa
σ'maxa  σ'mina
σ'mina  0  MPa
σ'aa  4.945 MPa
2
σ'maxa  σ'mina
σ'ma  4.945 MPa
2
σ'maxb 
σ'ab 
3  τa
3  τb
σ'maxb  39.664 MPa
σ'maxb  σ'minb
σ'minb  0  MPa
σ'ab  19.832 MPa
2
σ'maxb  σ'minb
σ'mb  19.832 MPa
2
S'e  0.5 S ut
S'e  275 MPa
3.
Calculate the unmodified endurance limit.
4.
Calculate the endurance limit modification factors for a nonrotating round pin.
Load
Cload  1
Size
A95 
π d
2
A95  78.54 mm
4
d equiv 
A95
d equiv  32.021 mm
0.0766
 d equiv 
Csize  1.189  

 mm 
Surface
A  4.51
Csurf
 Sut 
 A  

 MPa 
2
 0.097
b  0.265
Csize  0.849
(machined)
b
Csurf  0.847
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P0622.xmcd
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
5.
Temperature
Ctemp  1
Reliability
Creliab  0.659
6-22-2
(R = 99.999%)
Calculate the modified endurance limit.
S e  Cload  Csize Csurf  Ctemp Creliab S'e
6.
S e  130.42 MPa
Assuming a Case 3 load line, use equation (6.18e) to determine the factor of safety.
Part (a)
Nfa 
Part (b)
Nfb 
S e S ut
σ'aa S ut  σ'ma S e
S e S ut
σ'ab S ut  σ'mb S e
Nfa  21.3
Nfb  5.3
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-23a-1
PROBLEM 6-23a
Statement:
The beam in Figure P6-11a is subjected to a sinusoidal force-time function with Fmax = F and Fmin
= -F/2, where F and the beam's other data are given in row a of Table P6-5. Find the stress state in
the beam due to this loading and choose a material specification that will give a safety factor of 3
for N = 5E8 cycles.
Given:
Beam length
Distance to concen. load
Concentrated load
Moment of inertia
8
I  2.85 10
L
L  1  m
b  0.6 m
F  500  N
b
F
4
m
Distance to extreme fiber c  2.00 10
Design safety factor
Nd  3
2
m
8
Solution:
Cycle life
Nf  5  10
See Figure 6-23 and Mathcad file P0623a.
R2
R1
FIGURE 6-23
Free Body Diagram for Problem 6-23
1. The minimum, maximum, alternating, and mean components of the loads are:
Fmax  F
Fa 
2.
Fmax  Fmin
2

Mm  Fm b   1 

Fa  375  N
Fm 
σ'm 
b

Fmin  250  N
2
Fmax  Fmin
2
Fm  125  N
Ma  90 N  m
L
b

Mm  30 N  m
L
M a c
σ'a  63.158 MPa
I
Mm c
σ'm  21.053 MPa
I
Calculate the beam cross-section dimensions from I and c.
Beam depth
h  2  c
Beam width
w 
12 I
h
5.
F
Calculate the alternating and mean components of the maximum bending stress in the beam using equation
(4.11b). These are principal stresses and also von Mises stresses.
σ'a 
4.
Fmin 
Calculate the alternating and mean components of the maximum bending moment on the beam using the equati
in Figure B-2(a) in Appendix B.
Ma  Fa b   1 
3.
Fmax  500  N
3
h  40 mm
w  5.344  mm
Calculate the endurance limit modification factors for a nonrotating rectangular beam.
Load
Cload  1
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
A95  0.05 w h
Size
d equiv 
Surface
Csurf  1
Temperature
Ctemp  1
Reliability
Creliab  1
A95  10.688 mm
A95
2
d equiv  11.812 mm
0.0766
 d equiv 
Csize  1.189  

 mm 
6.
6-23a-2
 0.097
Csize  0.936
(R = 50%)
Determine the modified endurance limit as a function of the unknown endurance limit.
S e S ut  Cload  Csize Csurf  Ctemp Creliab 0.5 S ut
7.
Assuming a Case 3 load line, use equation (6.18e) as the design equation.
Nd =
8.
S e S ut  S ut
σ'a S ut  σ'm S e S ut
Solve the equations in steps 6 and 7 simultaneously for the desired S ut.
S ut 
Nd   2  σ'a  Cload  Csize Csurf  Ctemp Creliab σ'm
Cload  Csize Csurf  Ctemp Creliab
S ut  468  MPa
9.
Choose AISI 1020, cold-rolled steel (see Appendix A, Table A-9).
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-24a-1
PROBLEM 6-24a
Statement:
The beam in Figure P6-11b is subjected to a sinusoidal force-time function with Fmax = F and
Fmin = F/2, where F and the beam's other data are given in row a of Table P6-5. Find the stress
state in the beam due to this loading and choose a material specification that will give a safety
factor of 1.5 for N = 5E8 cycles.
Given:
Beam length
Concentrated load
Moment of inertia
L  1  m
F  500  N
8
I  2.85 10
4
m
Distance to extreme fiber c  2.00 10
Design safety factor
Nd  1.5
Nf  5  10
Cycle life
Solution:
L
F
2
m
M1
8
R1
See Figure 6-24 and Mathcad file P0624a.
FIGURE 6-24
1. The minimum, maximum, alternating, and mean components
of the loads are:
Fmax  F
Fa 
2.
3.
2
Fmax  500  N
Fmin 
Fa  125  N
Fm 
Ma  Fa L
Ma  125  N  m
Mm  Fm L
Mm  375  N  m
Fmin  250  N
2
Fmax  Fmin
2
Fm  375  N
Calculate the alternating and mean components of the maximum bending stress in the beam using equation
(4.11b). These are principal stresses and also von Mises stresses.
σ'm 
M a c
σ'a  87.719 MPa
I
Mm c
σ'm  263.158  MPa
I
Calculate the beam cross-section dimensions from I and c.
Beam depth
h  2  c
Beam width
w 
12 I
h
5.
F
Calculate the alternating and mean components of the maximum bending moment on the beam using the equati
in Figure B-1(a) in Appendix B.
σ'a 
4.
Fmax  Fmin
Free Body Diagram for Problem 6-24
3
h  40 mm
w  5.344  mm
Calculate the endurance limit modification factors for a nonrotating rectangular beam.
Load
Cload  1
Size
A95  0.05 w h
A95  10.688 mm
2
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
d equiv 
A95
6.
Csurf  1
Temperature
Ctemp  1
Reliability
Creliab  1
d equiv  11.812 mm
0.0766
 d equiv 
Csize  1.189  

 mm 
Surface
6-24a-2
 0.097
Csize  0.936
(R = 50%)
Determine the modified endurance limit as a function of the unknown endurance limit assuming the material is
steel.
S e S ut  Cload  Csize Csurf  Ctemp Creliab 0.5 S ut
7.
Assuming a Case 3 load line, use equation (6.18e) as the design equation.
Nd =
8.
S e S ut  S ut
σ'a S ut  σ'm S e S ut
Solve the equations in steps 6 and 7 simultaneously for the desired S ut.
S ut 
Nd   2  σ'a  Cload  Csize Csurf  Ctemp Creliab σ'm
Cload  Csize Csurf  Ctemp Creliab
S ut  676  MPa
9.
Choose AISI 1060 hot-rolled steel (see Appendix A, Table C-9).
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-25a-1
PROBLEM 6-25a
Statement:
The beam in Figure P6-11c is subjected to a sinusoidal force-time function with Fmax = F and
Fmin = 0, where F and the beam's other data are given in row a of Table P6-5. Find the stress
state in the beam due to this loading and choose a material specification that will give a safety
factor of 2.5 for N = 5E8 cycles.
Given:
Beam length
L  1  m
Distance between supports
b  0.6 m
Concentrated load
F  500  N
Moment of inertia
8
I  2.85 10
FIGURE 6-25
Free Body Diagram for Problem 6-25
Fmax  Fmin
2
Fmax  500  N
Fmin  0  N
Fa  250  N
Fm 
Ma  Fa ( b  L)
Ma  100  N  m
Mm  Fm ( b  L)
Mm  100  N  m
Fmin  0  N
Fmax  Fmin
2
Fm  250  N
Calculate the alternating and mean components of the maximum bending stress in the beam using equation
(4.11b). These are principal stresses and also von Mises stresses.
σ'm 
M a c
σ'a  70.175 MPa
I
Mm c
σ'm  70.175 MPa
I
Calculate the beam cross-section dimensions from I and c.
Beam depth
h  2  c
Beam width
w 
12 I
h
5.
8
R2
R1
Calculate the alternating and mean components of the maximum bending moment on the beam using the equati
in Figure B-1(a) in Appendix B.
σ'a 
4.
m
The minimum, maximum, alternating, and mean components of the loads are:
Fa 
3.
2
See Figure 6-25 and Mathcad file P0625a.
Fmax  F
2.
F
4
Nf  5  10
Cycle life
1.
b
m
Distance to extreme fiber c  2.00 10
Design safety factor
Nd  2.5
Solution:
L
3
h  40 mm
w  5.344  mm
Calculate the endurance limit modification factors for a nonrotating rectangular beam.
Load
Cload  1
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
A95  0.05 w h
Size
d equiv 
Surface
Csurf  1
Temperature
Ctemp  1
Reliability
Creliab  1
A95  10.688 mm
A95
2
d equiv  11.812 mm
0.0766
 d equiv 
Csize  1.189  

 mm 
6.
6-25a-2
 0.097
Csize  0.936
(R = 50%)
Determine the modified endurance limit as a function of the unknown endurance limit assuming the material is
steel.
S e S ut  Cload  Csize Csurf  Ctemp Creliab 0.5 S ut
7.
Assuming a Case 3 load line, use equation (6.18e) as the design equation.
Nd =
8.
S e S ut  S ut
σ'a S ut  σ'm S e S ut
Solve the equations in steps 6 and 7 simultaneously for the desired S ut.
S ut 
Nd   2  σ'a  Cload  Csize Csurf  Ctemp Creliab σ'm
Cload  Csize Csurf  Ctemp Creliab
S ut  550  MPa
9.
Choose AISI 1035 cold-rolled steel (see Appendix A, Table A-9).
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-26a-1
PROBLEM 6-26a
Statement:
Given:
The beam in Figure P6-11d is subjected to a sinusoidal force-time function with Fmax = F and Fmin
= -F, where F and the beam's other data are given in row a of Table P6-5. Find the stress state in
the beam due to this loading and choose a material specification that will give a safety factor of 6
for N = 5E8 cycles.
Beam length
L  1  m
Distance to concentrated load
a  0.4 m
Distance to 2nd support
b  0.6 m
Concentrated load
F  500  N
Moment of inertia
8
I  2.85 10
Distance to extreme fiber c  2.00 10
Design safety factor
Nfd  6
Solution:
1.
b
a
F
4
m
Nlife  5  10
Cycle life
L
2
R2
R1
R3
m
FIGURE 6-26A
8
Free Body Diagram for Problem 6-26
See Figures 6-26 and Mathcad file P0626a.
To determine the stresses, we must first get the maximum bending moment. From inspection of Figure P6-26,
write the load function equation
q(x) = R1<x>-1 - F<x - a>-1 + R2<x - b>-1 - R3<x - L>-1
2.
Integrate this equation from - to x to obtain shear, V(x)
V(x) = R1<x>0 - F<x - a>0 + R2<x - b>0 - R3<x - L>0
3.
Integrate this equation from - to x to obtain moment, M(x)
M(x) = R1<x>1 - F<x - a>1 + R2<x - b>1 - R3<x - L>1
4.
Integrate the moment function, multiplying by 1/EI, to get the slope.
(x) = [R1<x>2/2 - F<x - a>2/2 + R2<x - b>2/2 + R3<x - L>2/2 + C3]/EI
5.
Integrate again to get the deflection.
y(x) = [R1<x>3/6 - F<x - a>3/6 + R2<x - b>3/6 + R3<x - L>3/6 + C3x + C4]/EI
6.
Evaluate R1, R2, R3, C3 and C4
At x = 0, x = b, and x = L; y = 0, therefore, C4 = 0.
At x = L+, V = M = 0
R1  100  N
Guess
R2  100  N
2
R3  100  N
C3  5  N  m
Given
R1
6
R1
6
b 
F
3
F
3
L 
6
6
3
3
 ( b  a )  C 3 b = 0  N  m
3
 ( L  a) 
R2
6
3
3
 ( L  b )  C3 L = 0  N  m
R1  F  R2  R3 = 0  N
R1 L  F  ( L  a )  R2 ( L  b ) = 0  N  m
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-26a-2
 R1 
R 
 2   Find  R R R C 
1 2 3
3
 R3 
 
 C3 
R1  111.11 N
R2  472.22 N
2
R3  83.33  N
C3  5.556  N  m
x  0  in 0.002  L  L
7.
Define the range for x
8.
For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z,
and a value of one when it is greater than or equal to z.
S ( x z)  if ( x  z 1 0 )
9.
Write the shear and moment equations in Mathcad form, using the function S as a multiplying factor to get the
effect of the singularity functions.
V ( x)   R1 S ( x 0  in)  F  S ( x a )   R2 S ( x b )  R3 S ( x L)
M ( x)  R1 S ( x 0  in)  x  F  S ( x a )  ( x  a )  R2 S ( x b )  ( x  b )
10. Plot the shear and moment diagrams.
Shear Diagram
V ( x)
N
Moment Diagram
200
60
0
35
M ( x)
 200
10
Nm
 400
 15
 600
 40
0
200
400
FIGURE 6-26aB
600
800
3
1 10
0
200
400
600
x
x
mm
mm
800
3
1 10
Shear and Moment Diagrams for Problem 6-26a
11. From Figure 6-26aB, the maximum moment occurs at x = a. The maximum, minimum, alternating and mean
bending moments at x = a are:
Mmax  M ( a )
Ma 
Mm 
Mmax  Mmin
2
Mmax  Mmin
2
Mmax  44.4 N  m
Mmin  Mmax
Ma  44.444 N  m
Mm  0  N  m
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-26a-3
12. Calculate the alternating and mean components of the maximum bending stress in the beam using equation
(4.11b). These are principal stresses and also von Mises stresses.
σ'a 
M a c
σ'a  31.189 MPa
I
Mm c
σ'm 
σ'm  0  MPa
I
13. Calculate the beam cross-section dimensions from I and c.
Beam depth
h  2  c
Beam width
w 
h  40 mm
12 I
h
w  5.344  mm
3
14. Calculate the endurance limit modification factors for a nonrotating rectangular beam.
Load
Cload  1
Size
A95  0.05 w h
d equiv 
A95  10.688 mm
A95
d equiv  11.812 mm
0.0766
 d equiv 

 mm 
 0.097
Csize  1.189  
Surface
A  4.51
Temperature
Ctemp  1
Reliability
Creliab  1
2
Csize  0.936
b  0.265
 S ut 

 MPa 
Csurf  S ut  A  
b
(R = 50%)
15. Determine the modified endurance limit as a function of the unknown endurance limit.
S e S ut  Cload  Csize Csurf  S ut  Ctemp Creliab 0.5 S ut
16. Assuming a Case 3 load line, use equation (6.18e) as the design equation and solve for S ut.
Guess
S ut  100  MPa
Given
Nfd =
S e S ut  S ut
σ'a S ut  σ'm S e S ut
S ut  Find  S ut
S ut  447  MPa
Csurf  S ut  0.895
18. Choose AISI 1020, cold-rolled steel (see Appendix A, Table A-9).
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-27-1
PROBLEM 6-27
Statement:
A storage rack is to be designed to hold the paper roll of Problem 6-8 as shown in Figure P6-12.
Determine a suitable value for dimension a in the figure for an infinite-life fatigue safety factor of
2. Assume dimension b = 100 mm and that the mandrel is solid and inserts halfway into the
paper roll.
(a) The beam is a ductile material with S ut = 600 MPa.
(b) The beam is a cast-brittle material with S ut = 300 MPa.
Given:
Paper roll dimensions
Ductile tensile strength
3
OD  1.50 m
ID  0.22 m
Roll density
Design safety factor
ρ  984  kg m
Lroll  3.23 m
Brittle tensile strength
S utb  300  MPa
Nfd  2
S uta  600  MPa
W
Assumptions: The paper roll's weight creates a
concentrated load acting at the tip of
the mandrel. The mandrel's root fits
tightly in the stanchion so it can be
modeled as a cantilever beam. The
mandrel is machined, reliability is 90%,
and it operates at room temperature.
Solution:
1.
M1
Lm
R1
FIGURE 6-27
See Figure 6-27 and Mathcad file P0627.
Free Body Diagram used in Problem 6-27
Determine the weight of the roll and the length of the mandrel.
W 
Weight
π
4

2
2

 OD  ID  Lroll  ρ  g
W  53.9 kN
Lm  0.5 Lroll
Length
2.
a
Lm  1.615 m
The maximum moment occurs at a section where the mandrel root leaves the stanchion and is
Mmax  W  Lm
Mmax  87.04 kN  m
3.
The dynamic loading is repeated from 0 to Mmax on each stress cycle, thus Mmin  0  kN  m
4.
Part (a) - Calculate the alternating and mean components of the bending moment.
Ma 
Mm 
Mmax  Mmin
Ma  43520 N  m
2
Mmax  Mmin
Mm  43520 N  m
2
S'e  0.5 S uta
S'e  300 MPa
5.
Determine the unmodified endurance limit.
6.
Calculate the endurance limit modification factors for a nonrotating rectangular beam.
Load
Cload  1
Size
A95 ( a )  0.010462 a
2
 dequiv( a) 
Csize( a )  1.189  

 mm 
d equiv( a ) 
A95 ( a )
0.0766
 0.097
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
P0627.xmcd
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
A  4.51
Surface
Csurf
7.
6-27-2
b  0.265
b
 S uta 
 A  

 MPa 
Temperature
Ctemp  1
Reliability
Creliab  0.897
(machined)
Csurf  0.828
(R = 90%)
Calculate the modified endurance limit.
S e( a )  Cload  Csize( a )  Csurf  Ctemp Creliab S'e
8.
We can now determine the minimum required diameter, a. Using the distortion-energy failure theory with
the modified Goodman diagram, the bending stress will also be the only nonzero principal stress, which will
also be the von Mises stress. Assuming a Case 3 load line, use equation (6.18e) to determine the factor of
safety. Guess a  100  mm.
Bending stress
σ=
M c
I
Given
Nfd =
a 64
32 M
= M 
=
2
4
3
π a
π a
π a
3
32

S e( a )  S uta
Ma S uta  Mm S e( a )
a  Find ( a )
a  186.864 mm
a  190  mm
Round this up to the next higher even value
Using this value of a, the values of the functions of a are:
Csize( a )  0.787
S e( a )  175.371 MPa
The realized safety factor is
Nfa 
9.
π a
32
3

S e( a )  S uta
Nfa  2.1
Ma S uta  Mm S e( a )
S'e  0.4 S utb
Part (b) - Determine the unmodified endurance limit.
S'e  120 MPa
10. Calculate the endurance limit size modification factor for a nonrotating rectangular beam.
Size
A95 ( a )  0.010462 a
2
 dequiv( a) 
Csize( a )  1.189  

 mm 
d equiv( a ) 
A95 ( a )
0.0766
 0.097
11. Calculate the modified endurance limit.
S e( a )  Cload  Csize( a )  Csurf  Ctemp Creliab S'e
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-27-3
12. We can now determine the minimum required diameter, a. Using the distortion-energy failure theory with the
modified Goodman diagram, the bending stress will also be the only nonzero principal stress, which will also
be the von Mises stress. Assuming a Case 3 load line, use equation (6.18e) to determine the factor of safety.
Guess a  100  mm.
Bending stress
σ=
M c
I
Given
Nfd =
a 64
32 M
= M 
=
2
4
3
π a
π a
π a
32
3

S e( a )  S utb
Ma S utb  Mm S e( a )
a  Find ( a )
Round this up to the next higher even value
a  251.687 mm
a  252  mm
Using this value of a, the values of the functions of a are:
Csize( a )  0.766
S e( a )  68.253 MPa
The realized safety factor is
Nfb 
π a
32
3

S e( a )  S utb
Ma S utb  Mm S e( a )
Nfb  2.0
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-28-1
PROBLEM 6-28
Statement:
Figure P6-13 shows a forklift truck negotiating a 15 deg ramp to to drive onto a 4-ft-high loading
platform. The truck weighs 5 000 lb and has a 42-in wheelbase. Design two (one for each side)
1-ft-wide ramps of steel to have a safety factor of 2 for infinite life in the worst case of loading as
the truck travels up them. Minimize the weight of the ramps by using a sensible cross-sectional
geometry. Choose an appropriate steel or aluminum alloy.
Given:
Ramp angle
θ  15 deg
Platform height h  4  ft
Truck wheelbase Lt  42 in
Ramp width
Truck weight
w  12 in
W  5000 lbf
Assumptions: 1. The worst case is when the truck CG is located at the center of the beam's span.
2. Use a coordinate frame that has the x-axis along the long axis of the beam.
3. Ignore traction forces and the weight components along the x-axis of the beam.
4. There are two ramps, one for each side of the forklift.
See Figures 6-28 and Mathcad file P0628.
Solution:
L
b
a
CG a
y
CG b
R1

Fa
Wa
Fb
x
Wb
R2
FIGURE 6-28A
Dimensions and Free Body Diagram for Problem 6-28
1.
From Problem 3-28 the maximum bending moment in the ramp occurs at the rear wheel of the truck and is
Mmax  8324 ft  lbf
Mmax  99888 in lbf
Mmin  0  in lbf
The alternating and mean components of the bending moment are:
Ma 
Mm 
2.
Mmax  Mmin
2
Mmax  Mmin
2
Ma  49944 in lbf
Mm  49944 in lbf
The bending stress is the only stress component present and is, therefore, also the only nonzero principal
stress and is also the von Mises stress. The governing design equations then are
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P0628.xmcd
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
σ'a =
Ma
σ'm =
Z
6-28-2
Mm
Nfd =
Z
Combining these into a single equation
S e S ut
σ'a S ut  σ'm S e
Nfd =
Z  S e S ut
Ma S ut  Mm S e
3.
The approach will be to 1) choose a suitable factor of safety, 2) choose a suitable material and determine its
tensile strength and endurance limit, 3) from the equation above determine the required value of the section
modulus, 4) choose an appropriate cross-section for the ramp, and 5) determine the dimensions of the
cross-section.
4.
The following design choices have been made for this problem:
Design factor of safety
Nfd  2
Material
AISI 1095 steel, hot-rolled
Tensile strength
S ut  120  ksi
S'e  0.5 S ut
S'e  60 ksi
5.
Determine the unmodified endurance limit.
6.
Calculate the endurance limit modification factors for a nonrotating rectangular beam.
Load
Cload  1
Size
Csize  0.732
Surface
A  14.4
Csurf
7.
(initially guessed, and then found by iteration)
b  0.718
 S ut 
 A  

 ksi 
Temperature
Ctemp  1
Reliability
Creliab  0.897
(hot-rolled)
b
Csurf  0.463
(R = 90%)
Calculate the modified endurance limit.
S e  Cload  Csize Csurf  Ctemp Creliab S'e
8.
S e  18.237 ksi
Solve the design equation for the minimum section modulus, Z.
Z  Nfd 
Ma Sut  Mm Se
3
Z  6.309 in
S e S ut
This is the minimum allowable value of the section modulus.
9.
Assume a channel section such as that shown in Figure 6-28B. To keep it simple, let the thickness of the
flanges and web be the same. Choose 5/8-in thick plate, which is readily available. Then, t  0.625  in
10. The cross-sectional area of the ramp is A ( h )  w t  2  t ( h  t)
11. The distance to the CG is
cg( h ) 
1
A (h)
 w t 2

 2
2
 t h  t
2


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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
P0628.xmcd
mechanical,
photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-28-3
12. The moments of inertia of the web and a flange are
Iweb( h ) 
Ifl ( h ) 
w t
3
12
 w t  cg( h ) 

3
t ( h  t)
12
t


Flange
2
Web
2
h
 h  t  cg( h ) 
2

t
t
2


I ( h )  Iweb( h )  2  Ifl ( h )
h
13. The maximum stress will occur in the flange at the
top and is compressive. The distance from the
centroid up to the top of the flange is
w
c( h )  h  cg( h )
14. Using the known section modulus, solve for the
unknown flange height, h. Guess h  1  in
Given
Z=
I (h)
c( h )
h  Find ( h )
FIGURE 6-28B
Channel Section for Problem 6-28
h  4.304 in
Round this to
h  4.25 in
15. Check the size modification factor.
A95  0.05 w cg( h )  t ( h  cg( h ) )
d equiv 
A95
d equiv  5.858 in
0.0766
 d equiv 
Csize  0.869  

 in 
2
A95  2.628 in
 0.097
Csize  0.732
16. Summarizing, the ramp design dimensions are:
Width
w  12.00 in
Flange height
h  4.25 in
Shape
channel
Thickness
t  0.625 in
Material
1095 steel
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-29-1
PROBLEM 6-29
Statement:
A bar, 22 mm x 30 mm in cross-section, is loaded axially in tension with Fmin = -8 kN and Fmax =
+8 kN. A 10-mm hole passes through the center of the 30-mm side. Find the safety factor for
infinite life if the material has S ut = 500 MPa.
Given:
Bar width
w  30 mm
Maximum load
Fmax  8  kN
Bar thickness
h  22 mm
Minimum load
Fmin  8  kN
Hole diameter
Tensile strength
d  10 mm
S ut  500  MPa
Infinite life
F
Assumptions: Machined surfaces, temperature of 37C, and reliability
of 99.999%.
Solution:
1.
N =
See Figure 6-29 and Mathcad file P0629.
For completely reversed loading, the factor of safety is
Se
Kf  σ'athe uniform axial stress is the only stress component present,
Since
and
σ'a = σa
2.
N =
ø10
Calculate the endurance limit modification factors for an axial bar.
Load
Cload  0.7
(axial loading)
Size
Csize  1
(axial loading)
Surface
A  4.51
22
b  0.265
 Sut 
 A  

 MPa 
Temperature
Ctemp  1
Reliability
Creliab  0.659
b
Csurf  0.869
F
FIGURE 6-29
(R = 99.999%)
Free Body Diagram used in
Problem 6-29
Calculate the modified endurance limit.
S e  Cload  Csize Csurf  Ctemp Creliab S'e
5.
Kf  σa
S'e  250  MPa
Csurf
4.
Se
Calculate the unmodified endurance limit.
S'e  0.5 S ut
3.
30
S e  100.2  MPa
Determine the nominal (not increased by a stress concentration factor) alternating component of stress at the
hole.
Area
A  ( w  d )  h
Alternating load
Fa 
Alternating stress
σa 
Fmax  Fmin
2
Fa
A
A  440  mm
2
Fa  8  kN
σa  18.182 MPa
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6.
6-29-2
Determine the geometric stress concentration factor from Appendix C, Figure C-13.
7.
d
 7.9735 
d
2
3
d
 9.2659   

w
 w
w
4
5
d
d
 1.8145    2.9684  
w
 w
Kt  3.0039  3.753 
Determine the notch sensitivity of the material. Note from Figure 6-35 that the Neuber constant for steel in
tension is slightly lower that for torsional loading. However, comparison of values of a 1/2 obtained from the
dashed red curve with those in Table 6-6 indicates that, for tension as well as torsion, a value of 20 ksi should be
added to S ut to obtain a 1/2 from Table 6-6.
Lookup value of S ut
S'ut  S ut  20 ksi
Neuber constant
a  0.068  in
Notch radius
r  0.5 d
Notch sensitivity
q 
2
0.5
a  0.068  in
q  0.867
a
r
Determine the fatigue stress concentration factor from equation (6.11b).
Kf  1  q   Kt  1 
9.
S'ut  93 ksi
r  5  mm
1
1
8.
Kt  2.33
Kf  2.153
Determine the factor of safety against fatigue failure for the assumptions made.
Nf 
Se
Kf  σa
Nf  2.6
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-30-1
PROBLEM 6-30
Statement:
A bar, 22 mm x 30 mm in cross-section, is loaded axially in tension with Fmin = 0 kN and Fmax =
16 kN. A 10-mm hole passes through the center of the 30-mm side. Find the safety factor for
infinite life if the material has S ut = 500 MPa.
Given:
Bar width
w  30 mm
Maximum load
Fmax  16 kN
Bar thickness
h  22 mm
Minimum load
Fmin  0  kN
Hole diameter
Tensile strength
d  10 mm
S ut  500  MPa
Infinite life
Assumptions: Machined surfaces, temperature of 37C, and reliability
of 99.999%.
Solution:
See Figure 6-29 and Mathcad file P0629.
F
1.
For fluctuating loading, the factor of safety is
N =
S e S ut
Kf  σ'a S ut  Kfm σ'm S e
30
Since the uniform axial stress is the only stress component present,
σ'a = σa
2.
and
σ'm = σm
Calculate the unmodified endurance limit.
S'e  0.5 S ut
3.
S'e  250  MPa
Calculate the endurance limit modification factors for an axial bar.
Load
Cload  0.7
(axial loading)
Size
Csize  1
(axial loading)
Surface
A  4.51
 Sut 

 MPa 
b
Temperature
Ctemp  1
Reliability
Creliab  0.659
Csurf  0.869
F
FIGURE 6-30
(R = 99.999%)
Free Body Diagram used in
Problem 6-30
Calculate the modified endurance limit.
S e  Cload  Csize Csurf  Ctemp Creliab S'e
5.
22
b  0.265
Csurf  A  
4.
ø10
S e  100.2  MPa
Determine the nominal (not increased by a stress concentration factor) alternating and mean components of
stress at the hole.
Area
A  ( w  d )  h
A  440  mm
2
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
Alternating load
Fm 
Mean load
Alternating stress
σa 
σm 
Mean stress
6.
Fa 
6-30-2
Fmax  Fmin
Fa  8  kN
2
Fmax  Fmin
Fm  8  kN
2
Fa
σa  18.182 MPa
A
Fm
σm  18.182 MPa
A
Determine the geometric stress concentration factor from Appendix C, Figure C-13.
7.
d
 7.9735 
d
2
3
d
 9.2659   

w
 w
w
4
5
d
d
 1.8145    2.9684  
w
 w
Kt  3.0039  3.753 
Determine the notch sensitivity of the material. Note from Figure 6-35 that the Neuber constant for steel in
tension is slightly lower that for torsional loading. However, comparison of values of a 1/2 obtained from the
dashed red curve with those in Table 6-6 indicates that, for tension as well as torsion, a value of 20 ksi
should be added to S ut to obtain a 1/2 from Table 6-6.
Lookup value of S ut
S'ut  S ut  20 ksi
Neuber constant
a  0.068  in
Notch radius
r  0.5 d
Notch sensitivity
q 
S'ut  93 ksi
2
0.5
a  0.068  in
r  5  mm
1
1
8.
Kt  2.33
q  0.867
a
r
Determine the fatigue stress concentration factors from equations (6.11b) and (6.17).
Kf  1  q   Kt  1 
Kf  2.153
Kf   σm  σa  78 MPa
Assuming the yield strength for this material is about 400 MPa, we can use the first of equations (6.17) and
Kfm  Kf
9.
Kfm  2.153
Determine the factor of safety against fatigue failure for the assumptions made.
Nf 
S e S ut
Kf  σa S ut  Kfm σm S e
Nf  2.1
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-31-1
PROBLEM 6-31
Statement:
A bar, 22 mm x 30 mm in cross-section, is loaded axially in tension with Fmin = 8 kN and Fmax =
24 kN. A 10-mm hole passes through the center of the 30-mm side. Find the safety factor for
infinite life if the material has S ut = 500 MPa.
Given:
Bar width
w  30 mm
Maximum load
Fmax  24 kN
Bar thickness
h  22 mm
Minimum load
Fmin  8  kN
Hole diameter
Tensile strength
d  10 mm
S ut  500  MPa
Infinite life
Assumptions: Machined surfaces, temperature of 37C, and reliability
of 99.999%.
Solution:
1.
F
See Figure 6-29 and Mathcad file P0629.
For fluctuating loading, the factor of safety is
N =
S e S ut
Kf  σ'a S ut  Kfm σ'm S e
30
Since the uniform axial stress is the only stress component present,
and
σ'a = σa
2.
Calculate the unmodified endurance limit.
S'e  0.5 S ut
3.
ø10
σ'm = σm
S'e  250  MPa
Calculate the endurance limit modification factors for an axial bar.
Load
Cload  0.7
(axial loading)
Size
Csize  1
(axial loading)
Surface
A  4.51
22
b  0.265
 Sut 

 MPa 
b
Csurf  A  
Csurf  0.869
F
FIGURE 6-31
Temperature
Ctemp  1
Reliability
Creliab  0.659
Free Body Diagram used in
Problem 6-31
(R = 99.999%)
4. Calculate the modified endurance limit.
S e  Cload  Csize Csurf  Ctemp Creliab S'e
5.
S e  100.2  MPa
Determine the nominal (not increased by a stress concentration factor) alternating and mean components of
stress at the hole.
Area
A  ( w  d )  h
Alternating load
Fa 
Fmax  Fmin
2
A  440  mm
2
Fa  8  kN
© 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,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
Fm 
Mean load
Alternating stress
σm 
Mean stress
6.
σa 
6-31-2
Fmax  Fmin
Fa
σa  18.182 MPa
A
Fm
σm  36.364 MPa
A
Determine the geometric stress concentration factor from Appendix C, Figure C-13.
d
 7.9735 
d
2

 w
4
5
d
d
 1.8145    2.9684  
w
 w
Kt  3.0039  3.753 
7.
Fm  16 kN
2
w
 9.2659 
3
 
w
Kt  2.33
Determine the notch sensitivity of the material. Note from Figure 6-35 that the Neuber constant for steel in
tension is slightly lower that for torsional loading. However, comparison of values of a 1/2 obtained from the
dashed red curve with those in Table 6-6 indicates that, for tension as well as torsion, a value of 20 ksi should
be added to S ut to obtain a 1/2 from Table 6-6.
Lookup value of S ut
S'ut  S ut  20 ksi
Neuber constant
a  0.068  in
Notch radius
r  0.5 d
Notch sensitivity
q 
S'ut  93 ksi
2
0.5
a  0.068  in
r  5  mm
1
1
8.
d
q  0.867
a
r
Determine the fatigue stress concentration factors from equations (6.11b) and (6.17).
Kf  1  q   Kt  1 
Kf  2.153
Kf   σm  σa  117  MPa
Assuming the yield strength for this material is about 400 MPa, we can use the first of equations (6.17) and
Kfm  Kf
9.
Kfm  2.153
Determine the factor of safety against fatigue failure for the assumptions made.
Nf 
S e S ut
Kf  σa S ut  Kfm σm S e
Nf  1.8
© 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,
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Pearson Education, Inc., Upper Saddle River, NJ 07458.
MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-32-1
PROBLEM 6-32
Statement:
A bar, 22 mm x 30 mm in cross-section, is loaded axially in tension with Fmin = -4 kN and Fmax =
12 kN. A 10-mm hole passes through the center of the 30-mm side. Find the safety factor for
infinite life if the material has S ut = 500 MPa.
Given:
Bar width
w  30 mm
Maximum load
Fmax  12 kN
Bar thickness
h  22 mm
Minimum load
Fmin  4  kN
Hole diameter
Tensile strength
d  10 mm
S ut  500  MPa
Infinite life
Assumptions: Machined surfaces, temperature of 37C, and reliability
of 99.999%.
Solution:
1.
See Figure 6-29 and Mathcad file P0629.
F
For fluctuating loading, the factor of safety is
N =
S e S ut
Kf  σ'a S ut  Kfm σ'm S e
30
Since the uniform axial stress is the only stress component present,
and
σ'a = σa
2.
Calculate the unmodified endurance limit.
S'e  0.5 S ut
3.
S'e  250  MPa
Calculate the endurance limit modification factors for an axial bar.
Load
Size
Cload  0.7
Csize  1
Surface
A  4.51
b  0.265
 Sut 

 MPa 
b
Temperature
Ctemp  1
Reliability
Creliab  0.659
Csurf  0.869
F
FIGURE 6-32
(R = 99.999%)
Free Body Diagram used in
Problem 6-32
Calculate the modified endurance limit.
S e  Cload  Csize Csurf  Ctemp Creliab S'e
5.
22
(axial loading)
(axial loading)
Csurf  A  
4.
ø10
σ'm = σm
S e  100.2  MPa
Determine the nominal (not increased by a stress concentration factor) alternating and mean components of
stress at the hole.
Area
A  ( w  d )  h
Alternating load
Fa 
Fmax  Fmin
2
A  440  mm
2
Fa  8  kN
© 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.
MACHINE DESIGN - An Integrated Approach, 4th Ed.
Fm 
Mean load
Alternating stress
σm 
Mean stress
6.
σa 
6-32-2
Fmax  Fmin
Fm  4  kN
2
Fa
σa  18.182 MPa
A
Fm
σm  9.091  MPa
A
Determine the geometric stress concentration factor from Appendix C, Figure C-13.
7.
d
 7.9735 
d
2
3
d
 9.2659   

w
 w
w
4
5
d
d
 1.8145    2.9684  
w
 w
Kt  3.0039  3.753 
Determine the notch sensitivity of the material. Note from Figure 6-35 that the Neuber constant for steel in
tension is slightly lower that for torsional loading. However, comparison of values of a 1/2 obtained from the
dashed red curve with those in Table 6-6 indicates that, for tension as well as torsion, a value of 20 ksi should be
added to S ut to obtain a 1/2 from Table 6-6.
Lookup value of S ut
S'ut  S ut  20 ksi
Neuber constant
a  0.068  in
Notch radius
r  0.5 d
Notch sensitivity
q 
S'ut  93 ksi
2
0.5
a  0.068  in
r  5  mm
1
1
8.
Kt  2.33
q  0.867
a
r
Determine the fatigue stress concentration factors from equations (6.11b) and (6.17).
Kf  1  q   Kt  1 
Kf  2.153
Kf   σm  σa  59 MPa
Assuming the yield strength for this material is about 400 MPa, we can use the first of equations (6.17) and
Kfm  Kf
9.
Kfm  2.153
Determine the factor of safety against fatigue failure for the assumptions made.
Nf 
S e S ut
Kf  σa S ut  Kfm σm S e
Nf  2.3
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-33a-1
PROBLEM 6-33a
Statement:
For the bracket shown in Figure P6-14 subjected to a sinusoidal force-time function with Fmax = F
and Fmin = -F, where F and the beam's other data are given in row a of Table P6-6. Find the stress
states at points A and B due to this fully reversed loading and choose a ductile steel material
specification that will give a safety factor of 2 for infinite life. Assume a geometric
stress-concentration factor of 2.5 in bending and 2.8 in torsion.
Given:
Outside diameter
Geometric stress
od  20 mm
Kt  2.5
concentration factors
Kts  2.8
Design safety factor
Nd  2
A
B
T
Assumptions: The finish is machined, reliability is 50%,
and the bracket operates at room
temperature. The notch sensitivity q = 1
so that Kf = Kt.
Solution:
F
y
T
x
M
L
R
See Figure 6-33 and Mathcad file P0633a. FIGURE 6-33
Free Body Diagram of Tube for Problem 6-33
1.
From Problem 4-33a the stress components at point A are σx  8.38 MPa
2.
Calculate fatigue stresses and principal stresses.
Fatigue stresses
Principal stresses
τzx  16.76  MPa
σb  Kt σx
σb  20.95 MPa
τs  Kts τzx
τs  46.93 MPa
2
 σb 
2
σ1 
    τs
2
2
σb
σ1  58.56 MPa
σ2  0  MPa
σ3 
3.
σb
2
2

 σb 
2
   τs
2
 
σ3  37.61 MPa
Calculate the alternating von Mises effective stress (the mean component is zero).
σ'a 
2
2
σ1  σ1 σ3  σ3
σ'a  83.94 MPa
S'e S ut  0.5 S ut
4.
Calculate the unmodified endurance limit
5.
Determine the endurance limit modification factors
Load
Cload  1
Size
A95  0.010462 od
2
d eq 
A95
0.0766
(nonrotating round section)
d eq  7.391 mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-33a-2
 deq 
Csize  1.189  

 mm 
Surface
 0.097
Csize  0.979
A  4.51
b  0.265
 S ut 
Cs S ut  A  

 MPa 
b
(cold-drawn tubing)
Csurf  S ut  if  Cs S ut  1 1 Cs S ut 
6.
Temperature
Ctemp  1
Reliability
Creliab  1.0
(R = 50%)
Calculate the modified endurance limit
S e S ut  Cload  Csize Csurf  S ut  Ctemp Creliab S'e S ut
7.
Use the equation for the factor of safety for fully reversed loading to solve for S ut. Guess S ut  100  MPa
Given
8.
The varables that depend on S ut are:
Nd =
S e S ut
σ'a
Csurf  S ut  0.946
S ut  Find  S ut
S ut  362 MPa
S e S ut  167.88 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-34a-1
PROBLEM 6-34a
Statement:
For the bracket shown in Figure P6-14 subjected to a sinusoidal force-time function with Fmax = F
and Fmin = 0, where F and the beam's other data are given in row a of Table P6-6. Find the stress
states at points A and B due to this repeated loading and choose a ductile steel material
specification that will give a safety factor of 2 for infinite life. Assume a geometric
stress-concentration factor of 2.8 in bending and 3.2 in torsion.
Given:
Outside diameter
Geometric stress
od  20 mm
Kt  2.8
concentration factors
Kts  3.2
Design safety factor
Nd  2
A
B
T
Assumptions: The finish is machined, reliability is
50%, and the bracket operates at room
temperature. The notch sensitivity q =
1 so that Kf = Kt.
Solution:
F
y
T
x
M
L
R
See Figure 6-33 and Mathcad file P0633a. FIGURE 6-34
Free Body Diagram of Tube for Problem 6-34
1.
2.
From Problem 4-33a the stress components at point A are
σxmax  8.38 MPa
τzxmax  16.76  MPa
σxmin  0  MPa
τzxmin  0  MPa
Calculate the alternating and mean stress components.
σxa 
σxm 
τzxa 
τzxm 
3.
σxmax  σxmin
σxa  4.19 MPa
2
σxmax  σxmin
σxm  4.19 MPa
2
τzxmax  τzxmin
τzxa  8.38 MPa
2
τzxmax  τzxmin
τzxm  8.38 MPa
2
Calculate fatigue stresses and principal stresses.
Fatigue stresses
σba  Kt σxa
σba  11.73 MPa
τsa  Kts τzxa
τsa  26.82 MPa
σbm  Kt σxa
σbm  11.73 MPa
τsm  Kts τzxa
τsm  26.82 MPa
Principal stresses
2
 σba 
2
σ1a 
 
  τsa
2
 2 
σba
σ1a  33.32 MPa
σ2a  0  MPa
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obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-34a-2
2
 σba 
2
σ3a 
 
  τsa
2
2
 
σba
σ3a  21.58 MPa
2
 σbm 
2
σ1m 
 
  τsm
2
 2 
σbm
σ1m  33.32 MPa
σ2m  0  MPa
2
 σbm 
2
σ3m 
 
  τsm
2
 2 
σbm
4.
σ3m  21.58 MPa
Calculate the alternating and mean von Mises effective stress components.
2
2
σ'a 
σ1a  σ1a σ3a  σ3a
σ'm 
σ1m  σ1m σ3m  σ3m
2
σ'a  47.91 MPa
2
σ'm  47.91 MPa
S'e S ut  0.5 S ut
5.
Calculate the unmodified endurance limit
6.
Determine the endurance limit modification factors
Load
Cload  1
Size
A95  0.010462 od
2
d eq 
(nonrotating round section)
A95
d eq  7.391 mm
0.0766
 deq 
Csize  1.189  

 mm 
Surface
 0.097
Csize  0.979
A  4.51
b  0.265
 S ut 
Cs S ut  A  

 MPa 
b
(machined tubing)
Csurf  S ut  if  Cs S ut  1 1 Cs S ut 
7.
Temperature
Ctemp  1
Reliability
Creliab  1.0
(R = 50%)
Calculate the modified endurance limit
S e S ut  Cload  Csize Csurf  S ut  Ctemp Creliab S'e S ut
8.
Use the equation for the factor of safety for repeated loading assuming a Case 3 load line and using equation
(6.18e). Guess S ut  100  MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
Given
Nd =
S e S ut  S ut
σ'a S ut  σ'm S e S ut
S ut  Find  S ut
9.
The varables that depend on S ut are:
6-34a-3
Csurf  S ut  1
S ut  291 MPa
S e S ut  142.72 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-35a-1
PROBLEM 6-35a
Statement:
For the bracket shown in Figure P6-14 subjected to a sinusoidal force-time function with Fmax = F
and Fmin = -F, where F and the beam's other data are given in row a of Table P6-6. Find the stress
states at points A and B due to this fully reversed loading and choose a cast iron material
specification that will give a safety factor of 2 for infinite life. Assume a geometric
stress-concentration factor of 2.5 in bending and 2.8 in torsion.
Given:
Outside diameter
Geometric stress
od  20 mm
Kt  2.5
concentration factors
Kts  2.8
Design safety factor
Nd  2
A
B
T
Assumptions: The finish is as-cast, reliability is 50%,
and the bracket operates at room
temperature. There ia a fillet at the wall
with radius r  2  mm. However, set the
stress concentration factors and Csurf to 1
since cast iron's internal flaws mask these
effects.
Solution:
F
y
T
x
M
L
R
FIGURE 6-35
Free Body Diagram of Tube for Problem 6-35
See Figure 6-35 and Mathcad file P0635a.
1
From Problem 4-33a the stress components at point A are σx  8.38 MPa
2.
Calculate fatigue stress concentration factors and principal stresses. Let
Fatigue stresses
τzx  16.76  MPa
Kf  1
Kfs  1
σb  Kf  σx
σb  8.38 MPa
τs  Kfs τzx
τs  16.76 MPa
Principal stresses
σ1 
σb
2
2

 σb 
2
   τs
2
σ1  21.47 MPa
2
 σb 
2
σ3 
    τs
2
2
σb
σ3  13.09 MPa
σ2  0  MPa
3.
Calculate the alternating von Mises effective stress (the mean component is zero).
σ'a 
2
2
σ1  σ1 σ3  σ3
σ'a  30.21 MPa
S'e S ut  0.4 S ut
4.
Calculate the unmodified endurance limit
5.
Determine the endurance limit modification factors
Load
Cload  1
Size
A95  0.010462 od
2
(nonrotating round section)
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
d eq 
6-35a-2
A95
d eq  7.391 mm
0.0766
 deq 
Csize  1.189  

 mm 
6.
Surface
Csurf  1
Temperature
Ctemp  1
Reliability
Creliab  1.0
 0.097
Csize  0.979
( cast iron)
(R = 50%)
Calculate the modified endurance limit
S e S ut  Cload  Csize Csurf  Ctemp Creliab S'e S ut
7.
Use the equation for the factor of safety for fully reversed loading to solve for S ut. Guess S ut  100  MPa
Given
8.
The varables that depend on S ut are:
Nd =
S e S ut
σ'a
S ut  Find  S ut
S ut  154 MPa
S e S ut  60.43 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-36a-1
PROBLEM 6-36a
Statement:
For the bracket shown in Figure P6-14 subjected to a sinusoidal force-time function with Fmax = F
and Fmin = 0, where F and the beam's other data are given in row a of Table P6-6. Find the stress
states at points A and B due to this repeated loading and choose a cast iron material specification
that will give a safety factor of 2 for infinite life. Assume a geometric stress-concentration factor o
2.8 in bending and 3.2 in torsion.
Given:
Outside diameter
Geometric stress
od  20 mm
Kt  2.8
concentration factors
Kts  3.2
Design safety factor
Nd  2
Assumptions: The finish is as-cast, reliability is 50%,
and the bracket operates at room
temperature. There ia a fillet at the
wall with radius r  2  mm. However,
set the stress concentration factors and
Csurf to 1 since cast iron's internal flaws
mask these effects.
Solution:
1.
2.
Free Body Diagram of Tube for Problem 6-36
See Figure 6-36 and Mathcad file P0636a.
From Problem 4-33a the stress components at point A are
σxmax  8.38 MPa
τzxmax  16.76  MPa
σxmin  0  MPa
τzxmin  0  MPa
Calculate the alternating and mean stress components.
σxa 
σxm 
τzxa 
τzxm 
3.
FIGURE 6-36
σxmax  σxmin
σxa  4.19 MPa
2
σxmax  σxmin
σxm  4.19 MPa
2
τzxmax  τzxmin
τzxa  8.38 MPa
2
τzxmax  τzxmin
τzxm  8.38 MPa
2
Calculate fatigue stress concentration factors (set 1 for cast iron) and principal stresses
Fatigue stress
concentration
factors
Fatigue stresses
Kf  1
Kfs  1
Kfm  1
Kfsm  1
σba  Kf  σxa
σba  4.19 MPa
τsa  Kfs τzxa
τsa  8.38 MPa
σbm  Kfm σxa
σbm  4.19 MPa
τsm  Kfsm τzxa
τsm  8.38 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,
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-36a-2
Principal stresses
2
 σba 
2
σ1a 
 
  τsa
2
2
 
σba
σ1a  10.73 MPa
σ2a  0  MPa
2
 σba 
2
σ3a 
 
  τsa
2
2
 
σba
σ3a  6.54 MPa
2
 σbm 
2
σ1m 
 
  τsm
2
 2 
σbm
σ1m  10.73 MPa
σ2m  0  MPa
2
 σbm 
2
σ3m 
 
  τsm
2
 2 
σbm
4.
σ3m  6.54 MPa
Calculate the alternating von Mises effective stress (the mean component is zero).
2
2
σ'a 
σ1a  σ1a σ3a  σ3a
σ'm 
σ1m  σ1m σ3m  σ3m
2
σ'a  15.11 MPa
2
σ'm  15.11 MPa
S'e S ut  0.4 S ut
5.
Calculate the unmodified endurance limit
6.
Determine the endurance limit modification factors
Load
Cload  1
Size
A95  0.010462 od
2
d eq 
A95
d eq  7.391 mm
0.0766
 deq 
Csize  1.189  

 mm 
7.
(nonrotating round section)
Surface
Csurf  1
Temperature
Ctemp  1
Reliability
Creliab  1.0
 0.097
Csize  0.979
(cast iron)
(R = 50%)
Calculate the modified endurance limit
S e S ut  Cload  Csize Csurf  Ctemp Creliab S'e S ut
8.
Use the equation for the factor of safety for repeated loading assuming a Case 3 load line and using equation
(6.18e). Guess S ut  100  MPa
Given
Nd =
S e S ut  S ut
σ'a S ut  σ'm S e S ut
S ut  Find  S ut
S ut  107 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-37-1
PROBLEM 6-37
Statement:
A semicircular, curved beam as shown in Figure 5-37 has the dimensions given below. For a load
pair F = ±3 kN applied along the diameter, find the safety factor at the inner and outer fibers:
(a) If the beam is steel with S ut = 700 MPa,
(b) If the beam is cast-iron with S ut = 420 MPa.
Given:
(a) Tensile strength
S uta  700  MPa
(b) Tensile strength
S utb  420  MPa
Maximum load
Fmax  3  kN
Minimum load
Fmin  3  kN
Solution:
1.
See Figure 6-37 and Mathcad file P0637.
From Problem 4-37, the stresses at the inside radius are:
σi  409.9  MPa
Inside
These are based on a load of 14 kN. Since the stress in
a curved beam is directly proportional to the applied
load, we can determine the stresses at the inside surface
for this problem by applying the ratio 3/14 to this stress.
Thus,
σmax 
σmin 
3
14
3
14
 σi
σmax  87.836 MPa
 σi
σmin  87.836 MPa
FIGURE 6-37
These are the only stress components present on their
respective surfaces so they are also von Mises stresses.
2.
Free Body Diagrams for Problem 6-37
The dynamic loading in this problem is fully reversed. Determine the alternating stress component.
σ'a 
σmax  σmin
σ'a  87.836 MPa
2
Part (a)
S'ea  0.5 S uta
S'ea  350 MPa
3.
Calculate the unmodified endurance limit.
4.
Calculate the endurance limit modification factors for a nonrotating rectangular beam.
Load
Cload  0.7
(combined axial and bending loads)
Size
Section dims
w  25 mm
A95  0.05 w h
d equiv 
A95  31.25 mm
A95
 d equiv 

 mm 
A  4.51
2
d equiv  20.198 mm
0.0766
 0.097
Csize  1.189  
Surface
h  25 mm
b  0.265
Csize  0.888
(machined)
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
Csurf
5.
 S uta 
 A  

 MPa 
Temperature
Ctemp  1
Reliability
Creliab  0.897
6-37-2
b
Csurf  0.795
(R = 90%)
Calculate the modified endurance limit.
S ea  Cload  Csize Csurf  Ctemp Creliab S'ea
6.
S ea  155.15 MPa
Assuming no stress concentration, the fatigue factor of safety at the inner fiber for fully reversed loading is
Nfa 
S ea
Nfa  1.8
σ'a
Part (b)
S'eb  160  MPa
7.
Calculate the unmodified endurance limit using equation (6.5b).
8.
Calculate the endurance limit modification factors for a nonrotating rectangular beam.
Load
Cload  0.7
(combined axial and bending loads)
Size
Section dims
w  25 mm
A95  0.05 w h
d equiv 
A95  31.25 mm
A95
A  4.51
Csurf
9.
 0.097
Csize  0.888
b  0.265
 S utb 
 A  

 MPa 
Temperature
Ctemp  1
Reliability
Creliab  0.897
2
d equiv  20.198 mm
0.0766
 d equiv 
Csize  1.189  

 mm 
Surface
h  25 mm
(machined)
b
Csurf  0.910
(R = 90%)
Calculate the modified endurance limit.
S eb  Cload  Csize Csurf  Ctemp Creliab S'eb
S eb  81.21 MPa
10. Assuming no stress concentration, the fatigue factor of safety at the inner fiber for fully reversed loading is
Nfb 
S eb
σ'a
Nfb  0.92
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-38-1
PROBLEM 6-38
Statement:
A 42-mm-dia steel shaft with a 19-mm transverse hole is subjected to a sinusoidal combined
loading of  = ±100 MPa bending stress and steady torsion of 110 MPa. Find the safety factor
for infinite life if S ut = 1000 MPa.
Given:
Shaft diameter
D  42 mm
Max bending stress
σmax  100  MPa
Hole diameter
d  19 mm
Min bending stress
σmin  100  MPa
Steady torsion
τm  110  MPa
Tensile strength S ut  1000 MPa
Infinite life
Assumptions: Stresses given include stress concentration effects. Ground surfaces, temperature of 37C, and
reliability of 50%.
Solution:
1.
See Mathcad file P0638.
Calculate the alternating and mean von Mises stress components.
σ'a 
σmax  σmin
σ'm 
2.
σ'a  100 MPa
2
3  τm
σ'm  190.526 MPa
Calculate the unmodified endurance limit.
S'e  0.5 S ut
3.
S'e  500 MPa
Calculate the endurance limit modification factors for a rotating, round shaft.
Cload  1
Load
Size
Csize  1.189  
Surface
A  1.58
D


 mm 
Csurf
4.
(combined bending and torsion)
 0.097
 Sut 
 A  

 MPa 
Temperature
Ctemp  1
Reliability
Creliab  1
Csize  0.827
b  0.085
b
Csurf  0.878
(R = 50%)
Calculate the modified endurance limit.
S e  Cload  Csize Csurf  Ctemp Creliab S'e
5.
(ground)
S e  363.37 MPa
Assuming a Case 3 load line, determine the factor of safety against fatigue failure.
Nf 
S e S ut
σ'a S ut  σ'm S e
Nf  2.1
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-39-1
PROBLEM 6-39
Statement:
A 42-mm-dia steel shaft with a 19-mm transverse hole is subjected to a combined loading of  =
±100 MPa bending stress and an alternating torsion of ±110 MPa, which are 90 deg out of phase.
Find the safety factor for infinite life if S ut = 1000 MPa.
Given:
Shaft diameter
D  42 mm
Max bending stress
σmax  100  MPa
Hole diameter
d  19 mm
Min bending stress
σmin  100  MPa
Max torsional stress
τmax  110  MPa
Min torsional stress
τmin  110  MPa
Tensile strength S ut  1000 MPa
ϕ  90 deg
Phase angle
Assumptions: Stresses given include stress concentration effects. Ground surfaces, temperature of 37C, and
reliability of 50%.
Solution:
1.
See Mathcad file P0639.
The dynamic loading is fully reversed so both mean stresses are zero. Calculate the alternating SEQA stress
component using equation (6.23).
Q  2 
Stress ratio
τmax
Q  2.2
σmax
1
SEQAa 
2.
σmax 
3

4
2
1 
2
Q 
1
3
2
2
 Q  cos( 2  ϕ) 


SEQAa  190.526 MPa
S'e  500 MPa
Cload  1
(combined bending and torsion)


 mm 
Size
Csize  1.189  
Surface
A  1.58
D
 0.097
 Sut 

 MPa 
Temperature
Ctemp  1
Reliability
Creliab  1
Csize  0.827
b  0.085
(ground)
b
Csurf  A  
Csurf  0.878
(R = 50%)
Calculate the modified endurance limit.
S e  Cload  Csize Csurf  Ctemp Creliab S'e
5.
Q
2
Calculate the endurance limit modification factors for a rotating, round shaft.
Load
4.
16
4
Calculate the unmodified endurance limit.
S'e  0.5 S ut
3.
9
S e  363.37 MPa
Assuming a Case 3 load line, determine the factor of safety against fatigue failure.
Nf 
Se
SEQAa
Nf  1.9
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-40-1
PROBLEM 6-40
Statement:
Redesign the roll support of Problem 6-8 to be like that shown in Figure P6-16. The stub mandrels
insert to 10% of the roll length at each end. Design dimension a for an infinite-life factor of safety
of 2. See Problem 6-8 for additional data.
(a) The beam is a ductile material with S y = 450 MPa, S ut = 600 MPa
(b) The beam is a cast-brittle material with S ut = 300 MPa
Given:
Paper roll dimensions
OD  1.50 m
ID  0.22 m
3
ρ  984  kg m
Roll density
Design safety factor
Nfd  2
Lroll  3.23 m
Ductile tensile strength
S uta  600  MPa Brittle tensile strength
S utb  300  MPa
Assumptions: The paper roll's weight creates a concentrated load acting at the tip of the mandrel.
The mandrel's root fits tightly in the
stanchion so it can be modeled as a
cantilever beam. The mandrel is machined,
reliability is 99.999%, and it operates at
room temperature.
Solution:
1.
See Figure 6-40 and Mathcad file P0640.
Determine the weight of the roll, the load on each
support, and the length of the mandrel.
Weight of paper
W 

4
π
2
2
FIGURE 6-40

Free Body Diagram used in Problem 6-40
 OD  ID  Lroll  ρ  g
W  53.9 kN
2.
Load on one mandrel
F  0.5 W
F  26.95 kN
Length of mandrel
Lm  0.1 Lroll
Lm  0.323 m
The maximum moment occurs at a section where the mandrel root leaves the stanchion and is
Mmax  F  Lm
Mmax  8.704 kN  m
3.
The dynamic loading is repeated from 0 to Mmax on each stress cycle, thus Mmin  0  kN  m
4.
Part (a) - Calculate the alternating and mean components of the bending moment.
Ma 
Mm 
Mmax  Mmin
Ma  4352 N  m
2
Mmax  Mmin
Mm  4352 N  m
2
S'e  0.5 S uta
S'e  300 MPa
5.
Determine the unmodified endurance limit.
6.
Calculate the endurance limit modification factors for a nonrotating rectangular beam.
Load
Cload  1
Size
A95 ( a )  0.010462 a
2
d equiv( a ) 
A95 ( a )
0.0766
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-40-2
 dequiv( a) 
Csize( a )  1.189  

 mm 
A  4.51
Surface
Csurf
7.
 0.097
b  0.265
b
 S uta 
 A  

 MPa 
Temperature
Ctemp  1
Reliability
Creliab  0.659
(machined)
Csurf  0.828
(R = 99.999%)
Calculate the modified endurance limit.
S e( a )  Cload  Csize( a )  Csurf  Ctemp Creliab S'e
8.
We can now determine the minimum required diameter, a. Using the distortion-energy failure theory with the
modified Goodman diagram, the bending stress will also be the only nonzero principal stress, which will also
be the von Mises stress. Assuming a Case 3 load line, use equation (6.18e) to determine the factor of safety.
Guess a  100  mm.
Bending stress
σ=
M c
I
Given
Nfd =
a 64
32 M
= M 
=
2
4
3
π a
π a
π a
3
32

S e( a )  S uta
Ma S uta  Mm S e( a )
a  Find ( a )
a  92.421 mm
a  94 mm
Round this up to the next higher even value
Using this value of a, the values of the functions of a are:
Csize( a )  0.843
S e( a )  137.942 MPa
The realized safety factor is
Nfa 
9.
π a
32
3

S e( a )  S uta
Nfa  2.1
Ma S uta  Mm S e( a )
S'e  0.4 S utb
Part (b) - Determine the unmodified endurance limit.
S'e  120 MPa
10. Calculate the endurance limit size modification factor for a nonrotating rectangular beam.
Size
A95 ( a )  0.010462 a
2
 dequiv( a) 
Csize( a )  1.189  

 mm 
d equiv( a ) 
A95 ( a )
0.0766
 0.097
11. Calculate the modified endurance limit.
S e( a )  Cload  Csize( a )  Csurf  Ctemp Creliab S'e
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-40-3
12. We can now determine the minimum required diameter, a. Using the distortion-energy failure theory with
the modified Goodman diagram, the bending stress will also be the only nonzero principal stress, which will
also be the von Mises stress. Assuming a Case 3 load line, use equation (6.18e) to determine the factor of
safety. Guess a  100  mm.
Bending stress
σ=
M c
I
Given
Nfd =
a 64
32 M
= M 
=
2
4
3
π a
π a
π a
32
3

S e( a )  S utb
Ma S utb  Mm S e( a )
a  Find ( a )
Round this up to the next higher even value
a  124.874 mm
a  125  mm
Using this value of a, the values of the functions of a are:
Csize( a )  0.82
S e( a )  53.672 MPa
The realized safety factor is
Nfb 
π a
32
3

S e( a )  S utb
Ma S utb  Mm S e( a )
Nfb  2.0
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-41-1
PROBLEM 6-41
Statement:
A 10-mm ID steel tube carries liquid at 7 MPa. The pressure varies periodically from zero to
maximum. The steel has S ut = 400 MPa Determine the infinite-life fatigue safety factor for the
wall if its thickness is: a) 1 mm, b) 5 mm.
Given:
Tensile strength
S ut  400  MPa
Assumption: The tubing is long therefore the axial stress is zero. The finish is machined, reliability is 99.999%
and the tubing is at room temperature.
Solution:
See Mathcad file P0641.
t  1  mm
(a) Wall thickness is
1.
From Problem 4-41, this is a thick wall cylinder and the maximum principal stresses are:
σ1maxa  38.82  MPa
2.
σ2maxa  0  MPa
σ3maxa  7.00 MPa
Calculate the minimum, maximum, alternating, and mean von Mises effective stress using equation (5.7c).
σ'min  0  MPa
σ'maxa 
2
σ1maxa  σ1maxa σ3maxa  σ3maxa
σ'maxa  σ'min
σ'aa 
σ'maxa  σ'min
σ'ma  21.376 MPa
2
S'e  0.5 S ut
3.
Calculate the unmodified endurance limit.
4.
Calculate the endurance limit modification factors for axial loading.
Load
Size
Cload  0.7
Csize  1
Surface
A  4.51
Csurf
5.
σ'maxa  42.752 MPa
σ'aa  21.376 MPa
2
σ'ma 
2
S'e  200 MPa
(axial loading)
(axial loading)
b  0.265
 Sut 
 A  

 MPa 
Temperature
Ctemp  1
Reliability
Creliab  0.659
( machined )
b
Csurf  0.922
(R = 99.999%)
Calculate the modified endurance limit.
S e  Cload  Csize Csurf  Ctemp Creliab S'e
6.
S e  85.04 MPa
Assuming a Case 3 load line, the factor of safety from equation (6.18e) is
Na 
S e S ut
σ'aa S ut  σ'ma S e
Na  3.3
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
t  5  mm
(b) Wall thickness is
7.
From Problem 4-41, this is a thick wall cylinder and the principal stresses are:
σ1maxb  11.67  MPa
8.
σ2maxb  0  MPa
σ3maxb  7.00 MPa
Calculate the minimum, maximum, alternating, and mean von Mises effective stress using equation (5.7c).
σ'maxb 
2
σ1maxb  σ1maxb σ3maxb  σ3maxb
σ'maxb  σ'min
σ'ab 
2
σ'mb 
9.
6-41-2
σ'maxb  σ'min
2
2
σ'maxb  16.336 MPa
σ'ab  8.168 MPa
σ'mb  8.168 MPa
The endurance limit does not change from part a to b. Assuming a Case 3 load line, the factor of safety from
equation (6.18e) is
Nb 
S e S ut
σ'ab S ut  σ'mb S e
Nb  8.6
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-42-1
PROBLEM 6-42
Statement:
A cylindrical tank with hemispherical ends is required to hold 150 psi of pressurized air at room
temperature. The pressure cycles from zero to maximum. The steel has S ut = 500 MPa. Determine
the infinite-life fatigue safety factor if the tank diameter is 0.5 m with 1 mm wall thickness, and its
length is 1 m.
Given:
Tensile strength
Assumption:
The finish is machined, reliability is 99.999% and the tank is at room temperature.
Solution:
See Mathcad file P0642.
1.
S ut  500  MPa
From Problem 4-42, this is a thin wall cylinder and the maximum principal stresses are:
σ1max  259  MPa
2.
σ2max  129  MPa
σ3max  0  MPa
Calculate the minimum, maximum, alternating, and mean von Mises effective stress using equation (5.7c).
σ'min  0  MPa
σ'max 
σ'a 
2
σ1max  σ1max σ2max  σ2max
σ'max  σ'min
σ'max  224.301 MPa
σ'a  112.151 MPa
2
σ'max  σ'min
σ'm 
2
σ'm  112.151 MPa
2
S'e  0.5 S ut
3.
Calculate the unmodified endurance limit.
4.
Calculate the endurance limit modification factors for axial loading.
Load
Cload  0.7
(axial loading)
Size
Csize  1
(axial loading)
Surface
A  4.51
b  0.265
 Sut 

 MPa 
5.
Ctemp  1
Reliability
Creliab  0.659
( machined )
b
Csurf  A  
Temperature
S'e  250 MPa
Csurf  0.869
(R = 99.999%)
Calculate the modified endurance limit.
S e  Cload  Csize Csurf  Ctemp Creliab S'e
6.
S e  100.2 MPa
Assuming a Case 3 load line, the factor of safety from equation (6.18e) is
Nf 
S e S ut
σ'a S ut  σ'm S e
Nf  0.74
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-43-1
PROBLEM 6-43
Statement:
The paper rolls in Figure P6-17 are 0.9-m OD by 0.22-m ID by 3.23-m long and have a density of
984 kg/m3. The rolls are transfered from the machine conveyor (not shown) to the forklift truck
by the V-linkage of the off-load station, which is rotated through 90 deg by an air cylinder. The
paper then rolls onto the waiting forks of the truck. The forks are 38-mm thick by 100-mm wide
by 1.2-m long and are tipped at a 3-deg angle from the horizontal and have S ut = 600MPa. Find
the infinite-life safety factor for the two forks on the truck when the paper rolls onto it under two
different conditions (state all assumptions):
(a) The two forks are unsupported at their free end.
(b) The two forks are contacting the table at point A.
Given:
Tensile strength S ut  600  MPa
Fork width
Fork thickness
F
L fork
t
w  100  mm
t  38 mm
Assumptions: 1. The greatest bending moment will occur
when the paper roll is at the tip of the fork
for case (a) and when it is midway between
supports for case (b).
2. Each fork carries 1/2 the weight of a
paper roll.
3. For case (a), each fork acts as a cantilever
beam (see Appendix B-1(a)).
4. For case (b), each fork acts as a beam
that is built-in at one end and
simply-supported at the other.
5. The forks are machined, the reliability is
90%, and they operate at room temperature.
R1
M1
Case (a), Cantilever Beam
0.5 L fork
F
t
L fork
R1
R2
M2
Case (b), Fixed-Simply Supported Beam
Solution:
See Figure 6-43 and Mathcad file P0643.
FIGURE 6-43
1.
From Problem 4-43, the maximum stresses in the forks are:
Free Body Diagrams used in Problem 6-43
Case (a)
σmaxa  464.8  MPa
at the base of the fork.
Case (b)
σmaxb  87.2 MPa
also at the base of the fork.
Both cases
σmin  0  MPa
Since there are no other stress components present, these are also the maximum principal stresses and the von
Mises stresses. This is a repeated load problem.
Case (a)
2.
The dynamic loading is repeated from 0 to 1 for each paper roll that is transfered. The alternating and mean
components of the von Mises stress are:
Alternating von Mises stress
σ'a  0.5  σmaxa  σmin
σ'a  232.4  MPa
Mean von Mises stress
σ'm  0.5  σmaxa  σmin
σ'm  232.4  MPa
S'e  0.5 S ut
S'e  300  MPa
3.
Calculate the unmodified endurance limit.
4.
Calculate the endurance limit modification factors for a nonrotating rectangular beam.
Load
Cload  1
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
Size
6-43-2
A95  0.05 w t
d equiv 
 d equiv 
Csize  0.869  

 in 
Surface
A  4.51
Csurf
5.
Temperature
Ctemp  1
Reliability
Creliab  0.897
Csize  0.814
(machined)
b
Csurf  0.828
(R = 90%)
Calculate the modified endurance limit.
S e  Cload  Csize Csurf  Ctemp Creliab S'e
6.
0.0766
 0.097
b  0.265
 Sut 
 A  

 MPa 
A95
S e  181.357  MPa
Assuming a Case 3 load line, use equation (6.18e) to determine the factor of safety.
Case a
Nfa 
S e S ut
σ'a S ut  σ'm S e
Nfa  0.60
Case (b)
7.
8.
The dynamic loading is repeated from 0 to 1 for each paper roll that is transfered. The alternating and mean
components of the von Mises stress are:
Alternating von Mises stress
σ'a  0.5  σmaxb  σmin
σ'a  43.6 MPa
Mean von Mises stress
σ'm  0.5  σmaxb  σmin
σ'm  43.6 MPa
Assuming a Case 3 load line, use equation (6.18e) to determine the factor of safety.
Case a
Nfb 
S e S ut
σ'a S ut  σ'm S e
Nfb  3.2
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-44-1
PROBLEM 6-44
Statement:
Determine a suitable thickness for the V-links of the off-loading station of Figure P6-17 to limit th
deflections at the tips to 10 mm in any position during their rotation. Two V-links support the roll
at the 1/4 and 3/4 points along the roll's length and that each of the V-links is 10 cm wide by 1 m
long. What is the infinite-life safety factor when designed to limit deflection as above? S ut = 600
MPa. See Problem 4-43 for more information.
Given:
Roll OD
OD  0.90 m
Arm width
wa  100  mm
Roll ID
ID  0.22 m
Arm length
La  1000 mm
Roll length
Lroll  3.23 m
Max tip deflection
δtip  10 mm
Roll density
ρ  984  kg m
Mod of elasticity
Tensile strength
E  207  GPa
S ut  600  MPa
3
Assumptions: 1. The maximum deflection on an arm will occur just after it begins the transfer and just before it
completes it, i.e., when the angle is either zero or 90 deg., but after the tip is no longer supported b
the base unit.
2. At that time the roll is in contact with both arms ("seated" in the V) and will remain in that state
throughout the motion. When the roll is in any other position on an arm the tip will be supported.
3. The arm can be treated as a cantilever beam with nonend load.
4. A single arm will never carry more than half the weight of a roll.
5. The pipe to which the arms are attached has OD = 160 mm.
6. The V-links are machined, reliability is 90%, and they operate at room temperature.
Solution:
1.
See Figure 6-44 and Mathcad file P0644.
Determine the weight of the roll and the load on each
V-arm.
W 

4
π
2
2

 OD  ID  Lroll  ρ  g
F  0.5 W
2.
W  18.64  kN
F  9.32 kN
From Appendix B, Figure B-1, the tip deflection of
a cantilever beam with a concentrated load
located at a distance a from the support is
ymax =
F a
2
6  E I
 ( a  3  L)
where L is the beam length and I is the
cross-section moment of inertia. In this case
3
I=
3.
w a t a
12
Setting ymax = δtip and a  370  mm,
FIGURE 6-44
Free Body Diagram used in Problem 6-44
substituting for I and solving for ta
1
 2 F  a2  3 La  a 
ta 


E δtip  wa


Let the arm thickness be
3
ta  31.889 mm
ta  32 mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
4.
6-44-2
The maximum bending stress in the arm will be at its base where it joins the 160-mm-dia pipe. The bending
moment, moment of inertia, and distance to the outside fiber at that point are:
Mmax  F  a
Bending moment
Mmax  3.449  kN  m
Mmin  0  kN  m
Ma 
1
Mm 
1
2
  Mmax  Mmin
Ma  1.725  kN  m
  Mmax  Mmin
Mm  1.725  kN  m
2
c  0.5 t a
Distance to n.a.
I 
Moment of inertia
wa ta
5
I  2.731  10  mm
M a c
I
Mm c
σmnom 
5.
3
12
σanom 
Nom tensile stress
c  16 mm
I
4
σanom  101  MPa
σmnom  101  MPa
Determine the stress concentration factors. Figure E-10 comes the closest to our situation. Assuming that the
effective D/d-ratio is 2 and r/d is about 0.25, Kt  1.4. For a material with
3
ksi  10  psi
2
S ut  87 ksi
a  0.073  in
and, for the assumed value of r/d,
r  0.25 ta
r  8  mm
The notch sensitivity factor is
q 
1
1
q  0.885
a
r
and the fatigue stress concentration factor is
Kf  1  q   Kt  1 
6.
Kf  1.35
Assuming that Kfm  Kf , the actual alternating and mean components of stress at the point where the V-link
meets the central hub are
σa  Kf  σanom
σa  136.8  MPa
σm  Kfm σmnom
σm  136.8  MPa
7.
Since there are no other nonzero stress components at this point on the top of the arm, the von Mises stresses
are
and
σ'a  σa
σ'm  σm
8.
Determine the modified material strength.
Unmodified endurance limit
S'e  0.5 S ut
Load
Cload  1
Size
A95  0.05 wa ta
S'e  300  MPa
A95  160  mm
2
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
d eq 
A95
9.
d eq  45.703 mm
0.0766
 deq 
Csize  1.189  

 mm 
 0.097
 Sut 

 MPa 
Surface
Csurf  4.51 
Temperature
Ctemp  1
Reliability
Creliab  0.897
Endurance limit
6-44-3
S e  Cload  Csize Csurf  Ctemp Creliab S'e
Csize  0.821
 0.265
Csurf  0.828
(R = 90%)
S e  182.8  MPa
Calculate the factor of safety. Using the distortion energy theory and the modified Goodman theory, the
fatigue factor of safety for a V-link thickness of ta  32 mm is
Nf 
S ut S e
σ'a S ut  σ'm S e
Nf  1.0
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-45-1
PROBLEM 6-45
Statement:
Determine the infinite-life fatigue safety factor based on the tension load on the air-cylinder rod
in Figure P6-17. The tension load cycles from zero to maximum (compression loads below the
critical buckling load will not affect the fatigue life). The crank arm that it rotates is 0.3 m long
and the rod has a maximum extension of 0.5 m. The 25-mm-dia rod is solid steel with S ut = 600
MPa. State all assumptions.
Given:
Paper roll dimensions
OD  0.90 m
ID  0.22 m
Rod diameter
Tensile strength
d  25 mm
S ut  600  MPa
Lroll  3.23 m
3
ρ  984  kg m
Roll density
Assumptions: 1. The maximum force in the cylinder rod occurs when the transfer starts.
2. The cylinder and rod make an angle of 5.5 deg to the horizontal at the end of transfer.
3. The crank arm is 300 mm long and is 45 deg from vertical at the end of transfer.
4. The finish is machined, reliability is 90%, and the cylinder operates at room temperature.
5. The cylinder rod is fully retracted at the start of the transfer. At the end of the transfer
it will have extended 500 mm from its initial position.
Solution:
1.
See Figure 6-45 and Mathcad file P0645.
Determine the weight of the roll on the
forks.
W 

4
π
2
2
y

 OD  ID  Lroll  ρ  g
W  18.64 kN
2.
From the assumptions and Figure 6-45,
the x and y distances from the origin to
point A are,
Rax  300  cos( 45 deg)  mm
450.0
Ray  300  sin( 45 deg)  mm
W
Rx
Rax  212.132 mm
x
212.1
Ray  212.132 mm
Ry
A
F
3.
From Figure 6-45, the x distance from the
origin to point where W is applied is,
Rwx 
4.
OD
2
5.5°
212.1
Rwx  450 mm
Sum moments about the pivot point and
solve for the tensile force in the cylinder
rod.
FIGURE 6-45
Free Body Diagram at End of Transfer for V-link of Problem 6-45
W  Rwx  Fmax Rax sin( 5.5 deg)  Fmax Ray cos( 5.5 deg) = 0
Fmax 
5.
W  Rwx
Ray cos( 8  deg)  Rax sin( 8  deg)
Assume that the dynamic load is repeated so
Fmax  35.017 kN
tension
Fmin  0  kN
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6.
6-45-2
Determine the alternating and mean components of axial stress in the rod.
π d
A 
Area
Fm 
Mean load
σa 
Alternating stress
σm 
Mean stress
A  490.874 mm
4
Fa 
Alternating load
2
Fmax  Fmin
Fa  17.508 kN
2
Fmax  Fmin
Fm  17.508 kN
2
Fa
σa  35.668 MPa
A
Fm
σm  35.668 MPa
A
Nf =
S e S ut
7.
For fluctuating loading, the factor of safety is
8.
Since the uniform axial stress is the only stress component present,
and
σ'a = σa
9.
2
σ'a S ut  σ'm S e
σ'm = σm
Calculate the unmodified endurance limit.
S'e  0.5 S ut
S'e  300 MPa
10. Calculate the endurance limit modification factors for an axial bar.
Load
Cload  0.7
(axial loading)
Size
Csize  1
(axial loading)
Surface
A  4.51
Csurf
b  0.265
 Sut 
 A  

 MPa 
Temperature
Ctemp  1
Reliability
Creliab  0.897
(machined)
b
Csurf  0.828
(R = 90%)
11. Calculate the modified endurance limit.
S e  Cload  Csize Csurf  Ctemp Creliab S'e
S e  155.95 MPa
12. Determine the factor of safety against fatigue failure for the assumptions made.
Nf 
S e S ut
σa S ut  σm S e
Nf  3.5
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-46-1
PROBLEM 6-46
Statement:
The V-links of Figure P6-17 are rotated by the crank arm through a shaft that is 60 mm dia by 3.23
m long. Determine the maximum torque applied to this shaft during motion of the V-linkage and
find the infinite-life fatigue safety factor for the shaft if its S ut = 600 MPa. See Problem 6-43 for
more information.
y
Given:
Tensile strength S ut  600  MPa
Shaft diameter
d  60 mm
Assumptions: 1. The greatest torque will occur
when the link is horizontal and the
paper roll is located as shown in
Figure P6-17 or Figure 6-46.
2. The V-links are machined, use a
reliability of 90%, and operate at
room temperature.
Solution:
1.
See Figure 6-46 and Mathcad file
P0646.
W
From Problem 4-46, the maximum torsional
stress in the shaft is
T
τmax  197.88 MPa
Ry
60-mm-dia shaft
2.
3.
Although not exactly true, assume that the
load is fully reversed, then the minimum
torque is
τmin  197.88 MPa
FIGURE 6-46
Free Body Diagram used in Problem 6-46
Calculate the alternating component of the torsional stress in the shaft.
Alternating stress
4.
450.0
τa 
τmax  τmin
τa  197.88 MPa
2
Convert this to the corresponding component of the von Mises stress.
Alternating stress
σ'a 
3  τa
σ'a  342.738 MPa
S'e  0.5 S ut
5.
Calculate the unmodified endurance limit using equation (6.5a).
6.
Calculate the endurance limit modification factors for a solid, round steel shaft.
Load
Cload  1
Size
Csize  1.189  
Surface
A  4.51


 mm 
Csurf
d
 Sut 
 A  

 MPa 
Temperature
Ctemp  1
Reliability
Creliab  0.897
S'e  300 MPa
 0.097
b  0.265
Csize  0.799
(machined)
b
Csurf  0.828
(R = 90%)
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
7.
Calculate the modified endurance limit.
S e  Cload  Csize Csurf  Ctemp Creliab S'e
8.
6-46-2
S e  178.07 MPa
Calculate the factor of safety for the shaft.
Nf 
Se
σ'a
Nf  0.52
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-47-1
PROBLEM 6-47
Statement:
Determine the maximum forces on the pins at each end of the air cylinder of Figure P6-17.
Determine the infinite-life fatigue safety factor in these pins if they are 30-mm dia and in single
shear. S ut = 600 MPa. See Problem 6-43 for more information.
Given:
Pin diameter
d  30 mm
S ut  600  MPa
Tensile strength
Assumptions: 1. The maximum force in the cylinder rod occurs when the transfer starts.
2. The dynamic loading is fully reversed.
3. The finish is machined, reliability is 90%, and the pins are at room temperature.
Solution:
See Figure 6-47 and Mathcad file P0647.
y
W
Rx
x
212.1
Ry
A
F
8°
212.1
450.0
FIGURE 6-47
Free Body Diagram at Start of Transfer for V-link of Problem 6-47
τmax  65.7 MPa
1.
From Problem 4-47 the maximum shear stress on the pins is
2.
This is the only stress com- ponent so the alternating von Mises stress is σ'a 
3.
Calculate the unmodified endurance limit.
4.
Calculate the endurance limit modification factors for a nonrotating direct shear.
Size
Load
A95 
S'e  0.5 S ut
π d
S'e  300 MPa
2
A95  706.858 mm
4
2
Cload  1
d equiv 
A95
d equiv  96.062 mm
0.0766
 d equiv 
Csize  1.189  

 mm 
Surface
3  τmax σ'a  113.796 MPa
A  4.51
 0.097
b  0.265
Csize  0.764
(machined)
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P0647.xmcd
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
Csurf
5.
 Sut 
 A  

 MPa 
Temperature
Ctemp  1
Reliability
Creliab  0.897
6-47-2
b
Csurf  0.828
(R = 90%)
Calculate the modified endurance limit.
S e  Cload  Csize Csurf  Ctemp Creliab S'e
S e  170.12 MPa
6. Assuming a Case 3 load line, use equation (6.14) to determine the factor of safety.
Nf 
Se
σ'a
Nf  1.5
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-48-1
PROBLEM 6-48
Statement:
Figure P6-18 shows an exerciser for a 100-kg wheelchair racer. The wheelchair has 65-cm-dia
drive wheels separated by a 70-cm track width. Two free-turning rollers on bearings support the
rear wheels. The lateral movement of the chair is limited by the flanges. Design the 1-m-lomg
rollers as hollow tubes of aluminum (select alloy) to minimize the height of the platform and also
limit the roller deflections to 1 mm in the worst case. Specify suitable sized steel axles to support
the tubes on bearings. Calculate the fatigue safety factors at a life of 5E8 cycles.
Given:
Mass of chair
M  100  kg
Wheel diameter d w  650  mm
Track width
T  700  mm
Aluminum
Ea  71.7 GPa
Roller length
Lr  1000 mm
Steel
Es  207  GPa
Assumptions: 1. The CG of the chair with rider is
sufficiently close to the rear wheel that all
of the weight is taken by the two rear
wheels.
2. The small camber angle of the rear
wheels does not significantly affect the
magnitude of the forces on the rollers.
3. Both the aluminum roller and the steel
axle are simply supported. The steel axles
that support the aluminum tube are fixed
in the mounting block and do not rotate.
The aluminum tube is attached to them by
two bearings (one on each end of the
tubes, one for each axle). The bearings'
inner race is fixed, and the outer race
rotates with the aluminum tube. Each
steel axle is considered to be loaded as a
simply supported beam. Their diameter
must be less than the inner diameter of the
tubes to fit the roller bearings between
them.
4. All surfaces are machined, reliability is
90%, and parts are at room temperature.
Solution:
δ  1  mm
Maximum deflection
Modulus elasticity
W/2
F
F


FIGURE 6-48A
Free Body Diagram of One Wheel
used in Problem 6-48
See Figures 6-48 and Mathcad file P0648.
W  M  g
W  980.7  N
1.
Calculate the weight of the chair with rider. Weight of chair
2.
Calculate the forces exerted by the wheels on the rollers (see Figure 5-48A). From the FBD of a wheel,
summing vertical forces
2  F  cos( θ ) 
Let
θ  20 deg
W
2
=0
then
F 
W
4  cos( θ )
F  260.9  N
3.
The worst condition (highest moment at site of a stress concentration) will occur when the chair is all the way
to the left or right. Looking at the plane through the roller that includes the forces exerted by the wheels (the
FBD is shown in Figure 6-48B) the reactions R1 and R2 come from the bearings, which are inside the hollow
roller and are, themselves, supported by the steel axle.
4.
Solving for the reactions. Let the distance from R1 to F be a  15 mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
 M1
R2 Lr  F  ( a  T )  F  a = 0
 Fy
R1  2  F  R2 = 0
R2 
F  (2 a  T )
700
F
F
R2  190.5  N
Lr
15
R1  2  F  R2
5.
6-48-2
R2
R1
R1  331.3  N
The maximum bending moment will be at
the right-hand load and will be
1000
FIGURE 6-48B
Free Body Diagram of One Tube used in Problem 6-48
Mrmax  R2 Lr  ( a  T )
Mrmax  54.3 N  m
Note, if the chair were centered on the roller the maximum moment would be
Mc  F 
Lr  T
Mc  39.1 N  m
2
and this would be constant along the axle between the two loads, F.
6.
Note that the bearing positions are fixed regardless of the position of the chair on the roller.
Because of symmetry,
7.
Ra1  R1
Ra1  331.3  N
Ra2  R2
Ra2  190.5  N
1000
65
R1
R2
R a1
The maximum bending moment
occurs at R1 and is for
b  65 mm
R a2
1130
FIGURE 6-48C
Mamax  Ra1 b
Free Body Diagram of One Axle used in Problem 6-48
Mamax  21.5 N  m
8.
Determine a suitable axle diameter. Let the factor of safety against yielding in the axle be Nsa  3
9.
Tentatively choose a low-carbon steel for the axle, say AISI 1020, cold rolled with S y  393  MPa
10. At the top of the axle under the load R1 there is only a bending stress, which is also the von Mises stress. Set
this stress equal to the yield strength divided by the factor of safety.
σ' =
32 Mamax
π d a
3
=
Sy
Nsa
1
Solving for the axle diameter, d a
 32 Nsa Mamax 
d a  

π S y


Let the axle diameter be
d a  15 mm
3
d a  11.875 mm
made from cold-rolled AISI 1020 steel.
11. Suppose that bearing 6302 from Chapter 10, Figure 10-23, page 684 is used. It has a bore of 15 mm and an OD
of 42 mm. Thus, the inside diameter of the roller away from the bearings where the moment is a maximum will
be d i  40 mm. This will provide a 1-mm shoulder for axial location of the bearings.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
12. Design for a factor of safety of
Nra  3. Tentatively choose
6-48-3
150
700
F
2024-T4 aluminum with
F
S ut  440  MPa
and
S'e5E8  138  MPa
F
15
F
1000
13. A point on the outside diameter of the
roller will see completely reversed
bending, which will also be the only
nonzero principal stress. Thus,
σx =
Kf  Mrmax
Z
= σ' =
FIGURE 6-48D
Free Body Diagram of Roller with Chair in the Center.
Se
Nfr
where Kf is the fatigue stress concentration at the shoulder and S e is the modified endurance limit.
14. Tentatively choose (these values arrived at by iteration):
Outside diameter
d o  45 mm
Shoulder diameter
D  54 mm
Fillet radius
r  5  mm
15. Determine the fatigue stress concentration factor. From Figure E-2 and Table 6-6 for
r
do
D
 0.111
do
Kt  0.97098  


d
 o
r
 1.2
 0.21796
Kt  1.57
4
S ut  6.38  10  psi
1
q 
1
0.102
q  0.813
r
in
Kf  1  q   Kt  1 
Kf  1.46
16. Calculate the alternating von Mises stress component.
I 
π  4
4
  d o  d i 
64
c  0.5 d o
σ'a 
4
I  7.563  10  mm
4
c  22.5 mm
Kf  Mrmax  c
I
σ'a  23.6 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-48-4
17. Destermine the endurance limit at 5E8 cycles
Load factor
Cload  1
Size factor
 do 
Csize  1.189  

 mm 
 0.097
 Sut 
 4.51 

 MPa 
Surface factor
as machined
Csurf
Reliability
at 90%
Creliab  0.897
Csize  0.822
 0.265
Csurf  0.899
Modified endurance limit
S e5E8  Cload  Csize Csurf  Creliab S'e5E8
S e5E8  91.44  MPa
18. Determine the factor of safety for repeated loading.
Nfr 
S e5E8
Nfr  3.87
σ'a
19. The maximum deflection of the roller will occur when the chair is in the center of the roller. For this case the
reactions are both equal to the loads, F. Using Figure B-2(a) in Appendix B, the maximum deflection is at the
center of the roller and is for
a  150  mm
x  0.5 Lr
x  500  mm
E  71.7 GPa
ymax  2 
F
6  E I
  1 
a 3
3
  x  ( x  a ) 

L
r
 a
2
2

   a  3 a  Lr  2  Lr   x
L
 r





ymax  0.875  mm
This design meets the deflection requirement and has a reasonable factor of safety against fatigue failure while
allowing sufficient space for the bearings.
DESIGN SUMMARY
Axle
Roller
Material
AISI 1020 steel, cold-rolled
Material
2024-T4 aluminum
Diameter
d a  15 mm
Outside diameter
d o  45 mm
Length
1220 mm
Inside diameter
d i  40 mm
Shoulder dia
D  54 mm
Fillet radius
r  5  mm
Length
1040 mm
Center line spacing
c   d w  d o  sin( θ )
c  238  mm
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-49-1
PROBLEM 6-49
_____
Statement:
Figure P6-19 shows a machined pivot pin that is press-fit into part A and is slip fit in part B. If F
= 100 lb, l = 2 in, and d = 0.5 in, what is the pin's safety factor against fatigue when made of SAE
1020 cold-rolled steel? The loading is fully reversed and a reliability of 90% is desired. There is
a bending stress concentration factor Kt = 1.8 at the section where the pin leaves part A on the
right-hand side.
Given:
Applied force
F  100  lbf
Material strength S y  57 ksi
Total length, l
Pin dia
l  2.00 in
d  0.5 in
Beam length
S ut  68 ksi
L  0.5 l
Assumptions: 1. Since there is a slip fit between the pin and part B, part B offers no resistance to bending of
the pin and, since the pin is press-fit into part A, it can be modeled as a cantilever beam of
length l/2.
2. Part B distributes the concentrated force F so that, at the pin, it is uniformly distributed over
the exposed length of the pin.
Solution:
1.
See Mathcad file P0649.
Calculate the intensity of the uniformly distributed load acting over the length of the pin.
w 
2.
F
w  100.0 
L
lbf
in
A cantilever beam with uniform loading is shown in Figure B-1(b) in Appendix B. In this case, the dimension
a in the figure is zero. As shown in the figure, when a = 0, the maximum bending moment occurs at the
support and is
2
w L
Mmax 
3.
2
Calculate the moment of inertia and distance to the extreme fiber of the pin. The nominal alternating bending
stress in the beam is then found using equation 4.11b.
I 
π d
4
I  3.068  10
64
c  0.5 d
3
4
 in
c  0.250  in
Mmax c
σanom 
4.
Mmax  50.00  lbf  in
I
σanom  4074 psi
From Table 6-6, the Neuber constant for S ut  68 ksi is
1
2
a  0.096  in
5.
2
Using equation 6-13, the notch sensitivity for r  0.5 d is
q 
1
1
6.
a  0.096  in
q  0.839
a
r
The fatigue stress-concentration factor for Kt  1.8, from equation 6.11b, is
Kf  1  q   Kt  1 
Kf  1.67
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
7.
6-49-2
Because the stress state in the pin is simple, uniaxial stress, the alternating principal stress is equal to the
alternating tensile stress and is also equal to the alternating von Mises stress. Thus,
σa  Kf  σanom
σa  6.81 ksi
σ'  σa
S'e  0.5 S ut
S'e  34.0 ksi
8.
Calculate the unmodified endurance limit.
9.
Calculate the endurance limit modification factors for a non rotating round pin in bending.
Load
Cload  1
Size
A95  0.010462 d
d equiv 
2
A95
0.0766
A95  2.615  10
3
2
 in
d equiv  0.185  in
This is less than the lower limit in equation 6.7b, so use
Csize  1
(machined)
A  2.7
b  0.265
Surface
Csurf
 S ut 
 A  

 ksi 
Temperature
Ctemp  1
Reliability
Creliab  0.897
b
Csurf  0.883
(R = 90%)
10. Calculate the modified endurance limit.
S e  Cload  Csize Csurf  Ctemp Creliab S'e
S e  26.9 ksi
11. Using equation 6.14, calculate the factor of safety against a fatigue failure for this case of fully reversed
bending.
Nf 
Se
σ'
Nf  4.0
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-50-1
PROBLEM 6-50
_____
Statement:
Figure P6-19 shows a machined pivot pin that is press-fit into part A and is slip fit in part B. If F
= 100 N, l = 50 mm, and d = 16 mm, what is the pin's safety factor against fatigue when made of
class 50 cast iron? The loading is fully reversed and a reliability of 90% is desired. There is a
bending stress concentration factor Kt = 1.8 at the section where the pin leaves part A on the
right-hand side.
Given:
Applied force
F  100  N
Tensile strength S ut  359  MPa
Total length, l
l  50 mm
Beam length
L  0.5 l
Pin dia d  16 mm
Assumptions: 1. Since there is a slip fit between the pin and part B, part B offers no resistance to bending of
the pin and, since the pin is press-fit into part A, it can be modeled as a cantilever beam of
length l/2.
2. Part B distributes the concentrated force F so that, at the pin, it is uniformly distributed over
the exposed length of the pin.
Solution:
1.
See Mathcad file P0650.
Calculate the intensity of the uniformly distributed load acting over the length of the pin.
w 
2.
F
w  4.0
L
N
mm
A cantilever beam with uniform loading is shown in Figure B-1(b) in Appendix B. In this case, the dimension
a in the figure is zero. As shown in the figure, when a = 0, the maximum bending moment occurs at the
support and is
2
Mmax 
3.
w L
2
Calculate the moment of inertia and distance to the extreme fiber of the pin. The nominal alternating bending
stress in the beam is then found using equation 4.11b.
I 
π d
4
3
I  3.217  10  mm
64
c  0.5 d
σanom 
4.
Mmax  1250 N  mm
4
c  8.000  mm
Mmax c
I
σanom  3.108  MPa
Since this is a brittle material, so the full value of the geometric stress concentration factor Kt  1.8 will be
applied to the nominal stress using equation 4.31.
5.
Because the stress state in the pin is simple, uniaxial stress, the alternating principal stress is equal to the
alternating tensile stress and is also equal to the alternating von Mises stress. Thus,
σa  Kt σanom
σa  5.60 MPa
σ'  σa
S'e  0.5 S ut
S'e  179.5  MPa
6.
Calculate the unmodified endurance limit.
7.
Calculate the endurance limit modification factors for a non rotating round pin in bending.
Load
Cload  1
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
A95  0.010462 d
Size
d equiv 
A95
Csurf
2
d equiv  5.913  mm
0.0766
 Sut 
 A  

 MPa 
Temperature
Ctemp  1
Reliability
Creliab  0.897
b
Csurf  0.949
(R = 90%)
Calculate the modified endurance limit.
S e  Cload  Csize Csurf  Ctemp Creliab S'e
9.
A95  2.678  mm
This is less than the lower limit in equation 6.7b, so use
Csize  1
(machined)
A  4.51
b  0.265
Surface
8.
2
6-50-2
S e  152.7  MPa
Using equation 6.14, calculate the factor of safety against a fatigue failure for this case of fully reversed bending
Nf 
Se
σ'
Nf  27
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-51-1
PROBLEM 6-51
_____
Statement:
A component in the shape of a large sheet is to be fabricated from 7075-T651 aluminum, which
has a fracture toughness Kc = 24.2 MPa-m0.5 and a tensile yield strength of 495 MPa. Determine
the number of loading cycles that can be endured if the nominal stress varies from 0 to one half
the yield strength and the initial crack had a total length of 1.2 mm. The values of the coefficient
and exponent in equation 6.4 for this material are A = 5 x 10 -11 (mm/cyc) and n = 4.
Units:
cycle  1
Given:
Fracture toughness
Kc  24.2 MPa m
Yield strength
S y  495  MPa
Initial crack length
lo  1.2 mm
Coeff. and exponent
A  5  10
Solution:
1.
0.5
 11
 mm cycle
1
n  4
See Mathcad file P0651.
Calculate the minimum and maximum nominal stresses based on the yield strength and the stress level given in
the problem statement.
σmin  0  MPa
σmax 
2.
Sy
σmax  247.5 MPa
2
Determine the value of the geometry factor  from the discussion in Section 5.3 for a plate with a central crack.
β  1
3.
Using equation 5.14b, calculate the critical crack length for this material at the maximum stress condition.
 Kc 
a c   

π  β  σmax 
1
4.
2
a c  3.0 mm
Calculate the initial crack half-length.
a o  0.5 l o
5.
a o  0.60 mm
Using equations 6.3, write the stress intensity factor range as a function of crack half-length.
ΔK ( a )  β  π a   σmax  σmin
6.
Integrate equation 6.4 to find the number of cycles to failure.
1 
Nc   
A 




ac
1
 ΔK ( a ) 

0.5 
 MPa m 
4
da
5
Nc  7.2  10 cycle
ao
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-52-1
PROBLEM 6-52
_____
Statement:
A component in the shape of a large sheet is to be fabricated from 4340 steel, which has a fracture
toughness Kc = 98.9 MPa-m0.5. The sheets are inspected for crack flaws after fabrication, but the
inspection device cannot detect flaws smaller than 5 mm. Determine the minimum thickness
required for the sheet to have a minimum cycle life of 10 6 cycles (using fracture-mechanics criteria
if its width is 400 mm and the load normal to the crack varies from 20 to 170 kN. The values of th
coefficient and exponent in equation 6.4 for this material are A = 4 x 10 -9 (mm/cyc) and n = 3.
Units:
cycle  1
Given:
Fracture toughness
Kc  98.9 MPa m
Width of sheet
Load
W  400  mm
Fmin  20 kN
Coeff. and exponent
A  4  10
Cycles to failure
Nf  10  cycle
0.5
Fmax  170  kN
9
 mm cycle
The initial total crack length is lo  5  mm
Solution:
See Mathcad file P0652.
Write equations for the minimum and maximum nominal stresses as a function of the unknown thickness.
σmin( t) 
2.
n  3
6
Assumption:
1.
1
Fmin
σmax( t) 
W t
Fmax
W t
Determine the value of the geometry factor  from the discussion in Section 5.3 for a plate with a central crack.
β  1
3.
Using equation 5.14b, write an equation for the critical crack length as a function of t for this material at the
maximum stress condition.
 Kc 
a c( t)   

π  β  σmax( t) 
1
4.
2
Calculate the initial crack half-length.
a o  0.5 l o
5.
a o  2.50 mm
Using equations 6.3, write the stress intensity factor range as a function of crack half-length and sheet thickness
ΔK ( a t)  β  π a   σmax( t)  σmin( t) 
6.
Use equation 6.4 to find the minimum sheet thickness. First, guess a value: t  4  mm
Given

Nf  A = 





ac( t )
1
 ΔK ( a t) 
n
da
t  Find ( t)
t  3.2 mm

0.5 
 MPa m 
ao
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-53-1
PROBLEM 6-53
_____
Statement:
A closed, thin-wall cylinder is made from an aluminum alloy that has a fracture toughness of 38
MPa-m0.5 and has the following dimensions: length = 200 mm, OD = 84 mm, and ID = 70 mm. A
2.8-mm-deep semicircular crack is discovered on the inner diameter away from the ends, oriented
along a line parallel to the cylinder axis. If the cylinder is repeatedly pressurized from 0 to 75
MPa, how many pressure cycles can it withstand? The values of the coefficient and exponent in
equation 6.4 for this material are A = 5 x 10 -12 (mm/cyc) and n = 4. (Hint: the value of the
geometry factor for a semicircular surface flaw is  = 2/and the crack grows in the radial
direction).
Units:
cycle  1
Given:
Fracture toughness
0.5
Kc  38 MPa m
a o  2.8 mm
Initial crack depth
Solution:
1.
Cylinder dimensions
Internal pressure
L  200  mm
p min  0  MPa
Coeff. and exponent
Geometry factor
A  5  10
 mm cycle
β  0.6367
 12
1
n  4
See Mathcad file P0653.
Calculate the nominal hoop stress (tangential direction, normal to cylinder axis) using equation 4.49a based on
the pressure levels given in the problem statement.
r  0.5 ID
r  35.0 mm
σmin  p min 
r
t  0.5 ( OD  ID)
t  7.0 mm
σmin  0.0 MPa
t
σmax  p max 
2.
OD  84 mm
ID  70 mm
p max  75 MPa
r
σmax  375.0 MPa
t
Using equation 5.14b, calculate the critical crack length for this material at the maximum stress condition.
 Kc 
a c   

π  β  σmax 
1
2
a c  8.1 mm
However, since this is larger than the wall thickness, failure will occur when the crack reaches the OD so
a c  t
3.
a c  7.0 mm
Using equations 6.3, write the stress intensity factor range as a function of crack depth.
ΔK ( a )  β  π a   σmax  σmin
4.
Integrate equation 6.4 to find the number of cycles to failure.
1 
Nc   
A 




ac
1
 ΔK ( a ) 

0.5 
 MPa m 
4
da
6
Nc  1.34  10 cycle
ao
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-54-1
PROBLEM 6-54
_____
Statement:
A non rotating, hot-rolled, steel beam has a channel section with h = 64 mm and b = 127 mm. It is
loaded in repeated bending with the neutral axis through the web. Determine its corrected fatigue
strength with 90% reliability if it is used in an environment that has a temperature that is below
450C and has an ultimate tensile strength of 320 MPa.
Given:
Ultimate tensile strength
S ut  320  MPa
Reliability
Dimensions
R  0.90
h  64 mm
Solution:
1.
See Mathcad file P0654.
Calculate the uncorrected endurance limit using equation 6.5a.
S'e  0.5 S ut
2.
b  127  mm
S'e  160.0 MPa
Determine the loading factor from equation 6.7a.
Cload  1
3.
Determine the size factor from equations 6.7b and 6.7d, and Figure 6-25.
Area stressed above 95% of max
A95  0.05 b  h
Equivalent diameter
d equiv 
Size factor
4.
 d equiv 
Csize  1.189  

 mm 
A95  406.4 mm
A95
2
d equiv  72.8 mm
0.0766
 0.097
Csize  0.784
Determine the surface factor from equation 6.7e and Table 6-3.
From Table 6-3
Surface factor
A  57.7
Csurf
 Sut 
 A  

 MPa 
b  0.718
b
Csurf  0.917
5.
Determine the temperature factor from equation 6.7f. Since T < 450C, Ctemp  1.
6.
Determine the reliability factor from Table 6-4, Creliab  0.897.
7.
Using equation 6.6, calculate the corrected endurance limit.
S e  Cload  Csize Csurf  Ctemp Creliab S'e
S e  103.3 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-55-1
PROBLEM 6-55
_____
Statement:
A non rotating, machined, steel rod has a round section with d = 50 mm. It is loaded with a
fluctuating axial force. Determine its corrected fatigue strength with 99% reliability if it is used in
an environment that has a temperature below 450C and has an ultimate tensile strength of 480
MPa.
Given:
Ultimate tensile strength
S ut  480  MPa
Reliability
Dimensions
R  0.99
d  50 mm
Solution:
1.
See Mathcad file P0655.
Calculate the uncorrected endurance limit using equation 6.5a.
S'e  0.5 S ut
2.
S'e  240.0 MPa
Determine the loading factor from equation 6.7a.
Cload  0.70
3.
The size factor for an axially loaded member is Csize  1
4.
Determine the surface factor from equation 6.7e and Table 6-3.
From Table 6-3
Surface factor
A  4.51
Csurf
 Sut 
 A  

 MPa 
b  0.265
b
Csurf  0.878
5.
Determine the temperature factor from equation 6.7f. Since T < 450C, Ctemp  1.
6.
Determine the reliability factor from Table 6-4, Creliab  0.814.
7.
Using equation 6.6, calculate the corrected endurance limit.
S e  Cload  Csize Csurf  Ctemp Creliab S'e
S e  120.1 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-56-1
PROBLEM 6-56
_____
Statement:
A non rotating, cold-drawn, steel rod has a round section with d = 76 mm. It is loaded in
repeated torsion. Determine its corrected fatigue strength with 99% reliability if it is used in an
environment that has a temperature of 500C and has an ultimate tensile strength of 855 MPa.
Given:
Ultimate tensile strength
S ut  855  MPa
Reliability
Dimensions
Temperature
R  0.99
d  76 mm
T  500  C
Solution:
1.
See Mathcad file P0656.
Calculate the uncorrected endurance limit using equation 6.5a.
S'e  0.5 S ut
2.
S'e  427.5  MPa
Determine the loading factor from equation 6.7a.
Cload  1
3.
Determine the size factor from equation 6.7b.
Size factor
4.


 mm 
Csize  1.189  
 0.097
Csize  0.781
Determine the surface factor from equation 6.7e and Table 6-3.
A  4.51
From Table 6-3
Surface factor
5.
d
Csurf
b  0.265
 Sut 
 A  

 MPa 
b
Csurf  0.754
Determine the temperature factor from equation 6.7f.
Ctemp  1  0.0058 

T  450  C 
C


6.
Determine the reliability factor from Table 6-4, Creliab  0.814.
7.
Using equation 6.6, calculate the corrected endurance limit.
S e  Cload  Csize Csurf  Ctemp Creliab S'e
Ctemp  0.710
S e  145.5  MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-57-1
PROBLEM 6-57
_____
Statement:
A non rotating, ground, steel rod has a rectangular section with h = 60 mm and b = 40 mm. It is
loaded in repeated bending. Determine its corrected fatigue strength with 99.9% reliability if it is
used in an environment that has a temperature that is below 450C and has an ultimate tensile
strength of 1550 MPa.
Given:
Ultimate tensile strength
S ut  1550 MPa
Reliability
Dimensions
R  0.999
h  60 mm
Solution:
1.
b  40 mm
See Mathcad file P0657.
Calculate the uncorrected endurance limit using equation 6.5a (S ut exceeds 1400 MPa).
S'e  700  MPa
2.
Determine the loading factor from equation 6.7a.
Cload  1
3.
Determine the size factor from equations 6.7b and 6.7d, and Figure 6-25.
Area stressed above 95% of max
A95  0.05 b  h
Equivalent diameter
d equiv 
Size factor
4.
 d equiv 
Csize  1.189  

 mm 
A95  120.0 mm
A95
2
d equiv  39.6 mm
0.0766
 0.097
Csize  0.832
Determine the surface factor from equation 6.7e and Table 6-3.
From Table 6-3
Surface factor
A  1.58
Csurf
 Sut 
 A  

 MPa 
b  0.085
b
Csurf  0.846
5.
Determine the temperature factor from equation 6.7f. Since T < 450C, Ctemp  1.
6.
Determine the reliability factor from Table 6-4, Creliab  0.753.
7.
Using equation 6.6, calculate the corrected endurance limit.
S e  Cload  Csize Csurf  Ctemp Creliab S'e
S e  371.2 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-58-1
PROBLEM 6-58
_____
Statement:
A steel, grooved shaft similar to that shown in Figure C-5 (Appendix C) is to be loaded in bending
Its dimensions are: D = 57 mm, d = 38 mm, r = 3 mm. Determine the fatigue stress- concentration
factor if the material S ut = 1130 MPa.
Given:
Dimensions:
D  57 mm
Tensile strength S ut  1130 MPa
Solution:
See Figure C-5 and Mathcad file P0658.
1.
r  3  mm
The geometric stress-concentration factor is found from the equation in Figure C-5.
For
D
d
Kt  A  
 1.500
r
A  0.93894
b  0.32380
and
b

d
2.
d  38 mm
Kt  2.14
The Neuber constant is found by linear interpolation of the values in Table 6-6.
S ut  163.9  ksi
2
a 1  0.031  in
a 
3.
S ut  S 2
S1  S2
S 1  160  ksi
  a 1  a 2  a 2
2
a 2  0.024  in
S 2  180  ksi
0.5
a  0.030  in
Calculate the notch sensitivity using equation 6.13.
q 
1
1
q  0.920
a
r
4.
The fatigue stress-concentration factor can now be found from equation 6.11b.
Kf  1  q   Kt  1 
Kf  2.05
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-59-1
PROBLEM 6-59
_____
Statement:
A steel shaft with a transverse hole similar to that shown in Figure C-8 (Appendix C) is to be
loaded in torsion. Its dimensions are: D = 32 mm, d = 3 mm. Determine the fatigue stressconcentration factor if the material S ut = 808 MPa.
Given:
Dimensions:
Tensile strength
Solution:
See Figure C-8 and Mathcad file P0659.
1.
D  32 mm
S ut  808  MPa
d  3  mm
The geometric stress-concentration factor is found from an equation in Figure C-8. Although the maximum
torsional stress is on the surface, it will be almost a maximum just below the surface so it will be conservative
to use curve B.
2.
d
3
Kt  3.34
The Neuber constant is found by linear interpolation of the values in Table 6-6. However, since the loading is
torsional, 20 ksi must be added to the value of S ut that is used in the table (see the text in Figure 6-36, Part 1).
S utt  S ut  20 ksi
2
a 1  0.044  in
a 
3.
d
 27.159 
2
d
 30.231   

D
 D
 D
4
5
6
d
d
d
 393.19    650.39    15.451  
 D
 D
 D
Kt  3.9702  9.292 
S utt  S 2
S1  S2
S utt  137.2  ksi
S 1  130  ksi
  a 1  a 2  a 2
2
a 2  0.039  in
S 2  140  ksi
0.5
a  0.040  in
Calculate the notch sensitivity using equation 6.13. Let r  0.5 d
q 
1
1
q  0.857
a
r
4.
The fatigue stress-concentration factor can now be found from equation 6.11b.
Kf  1  q   Kt  1 
Kf  3.00
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-60-1
PROBLEM 6-60
_____
Statement:
A hardened aluminum filleted flat bar similar to that shown in Figure C-9 (Appendix C) is to be
loaded axially. Its dimensions are: D = 1.20 in, d = 1.00 in, r = 0.10 in. Determine the fatigue
stress-concentration factor if the material S ut = 76 ksi.
Given:
Dimensions:
Tensile strength
Solution:
See Figure C-9 and Mathcad file P0660.
1.
D
d
Kt  A  
 1.200
r
r  0.100  in
A  1.03510
b  0.25084
and
b

d
Kt  1.84
The Neuber constant is found by linear interpolation of the values in Table 6-8.
2
a 1  0.144  in
a 
3.
d  1.00 in
The geometric stress-concentration factor is found from the equation in Figure C-9.
For
2.
D  1.20 in
S ut  76 ksi
S ut  S 2
S1  S2
S 1  70 ksi
  a 1  a 2  a 2
2
a 2  0.131  in
S 2  80 ksi
0.5
a  0.136  in
Calculate the notch sensitivity using equation 6.13.
q 
1
1
q  0.699
a
r
4.
The fatigue stress-concentration factor can now be found from equation 6.11b.
Kf  1  q   Kt  1 
Kf  1.59
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-61-1
PROBLEM 6-61
Statement:
A rotating shaft with a shoulder fillet seated in the inner race of a rolling contact bearing with the
shoulder against the edge of the bearing is shown in Figure P6-20. The bearing has a slight
eccentricity that induces a fully reversed bending moment in the shaft as it rotates.
Measurements indicate that the resulting alternating stress amplitude due to bending is a = 57
MPa. The torque on the shaft fluctuates from a high of 90 N-m to a low of 12 N-m and is in
phase with the bending stress. The shaft is ground and its dimensions are: D = 23 mm, d = 19
mm, and r = 1.6 mm. The shaft material is SAE 1040 cold-rolled steel. Determine the infinite-life
fatigue safety factor for the shaft for a reliability of 99%.
Given:
Strength SAE 1040 CR
S ut  586  MPa Alternating bending stress σxa  57 MPa
Fluctuating torque
Tmax  90 N  m Tmin  12 N  m
Shaft dimensions
D  23 mm
Solution:
1.
Determine the mean and alternating components of the fluctuating torsional stress.
c 
J 
Polar moment of inertia
d
c  9.5 mm
2
π d
4
4
J  1.279  10  mm
32
τxymax 
Torsional stress
τxymin 
Tmax c
J
Tmin c
J
τxymax  τxymin
τxym 
2
τxymax  τxymin
τxya 
2
4
τxymax  66.827 MPa
τxymin  8.91 MPa
τxym  37.869 MPa
τxya  28.958 MPa
Using Appendix C, determine the geometric stress concentration factors for the bending and torsional stresses.
Bending (Fig. C-2): For
D
d
r
 1.211
Kt  A  
r
d
 0.084
Torsion (Fig. C-3): For
D
d
 1.211
Kts  A  
A  0.97098
b  0.21796
b

d
r

d
3.
r  1.6 mm
See Figure P6-20 and Mathcad file P0661.
Distance to outside fiber
2.
d  19 mm
Kt  1.665
r
d
 0.084
A  0.83425
b  0.21649
b
Kts  1.425
Calculate the notch sensitivity of the material for bending and torsion using Table 6-6.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-61-2
Bending:
2
Neuber constant (for S ut  586.0  MPa)
1
q b 
Notch sensitivity
a  0.075  in
q b  0.77
a
1
r
Torsion:
Neuber constant (for S ut  20 MPa  606.0  MPa)
1
q s 
Notch sensitivity
1
4.
5.
2
a  0.0585  in
q s  0.811
a
r
Calculate the fatigue stress concentration factors for bending and torsion using equation 6.11b.
Bending
Kf  1  q b  Kt  1 
Kf  1.512
Torsion
Kfs  1  q s  Kts  1 
Kfs  1.345
Determine what, if any, fatigue stress concentration factor should be applied to the mean torsional stress.
S y  490  MPa
Yield strength SAE 1040 CR
Evaluate
Kfs  2  τxymax  179.8  MPa
which is less than S y so
Kfsm  Kfs
6.
7.
Calculate the mean and alternating components of the stresses increased by the appropriate fatigue stress
concentration factors.
Bending
σm  0  MPa
σa  Kf  σxa
σa  86.188 MPa
Torsion
τm  Kfsm τxym
τm  50.934 MPa
τa  Kfs τxya
τa  38.949 MPa
Find the mean and alternating von Mises stresses using equations 6.22b (with y = 0).
2
Mean
σ'm 
3  τm
Alternating
σ'a 
σa  3  τa
2
σ'm  88.22  MPa
2
σ'a  109.451  MPa
8. Calculate the unmodified endurance limit.
S'e  0.5 S ut
S'e  293  MPa
9. Calculate the endurance limit modification factors for a rotating, round shaft.
Load
Cload  1
(combined bending and torsion)
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
Size
Csize  1.189  


mm
 
Surface
A  1.58
d
6-61-3
 0.097
Csize  0.894
b  0.085
 Sut 

 MPa 
b
Csurf  A  
Temperature
Ctemp  1
Reliability
Creliab  0.814
(ground)
Csurf  0.919
(R = 99%)
10. Calculate the modified endurance limit.
S e  Cload  Csize Csurf  Ctemp Creliab S'e
S e  196  MPa
11. Assuming a Case 2 load line, determine the factor of safety against fatigue failure.
Nf 
Se

1 
σ'a 
σ'm 
S ut


Nf  1.5
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-62-1
PROBLEM 6-62
Statement:
A tension member in a machine is filleted as shown in Figure P6-21. The member has a
manufacturing defect that causes the fluctuating tension load to be applied eccentrically
resulting in a fluctuating bending load as well. Measurements indicate that the maximum
bending stress is 16.4 MPa and the minimum is 4.1 MPa. The tensile load fluctuates from a high
of 3.6 kN to a low of 0.90 kN and is in phase with the bending stress. The member is machined
and its dimensions are: D = 33 mm, d = 25 mm, h = 3 mm and r = 3 mm. The material is SAE 1020
cold-rolled steel. Determine the infinite-life fatigue safety factor for the member for a reliability
of 90%.
Given:
Strength SAE 1020 CR
S ut  380  MPa σxmax  16.4 MPa
Fluctuating tension
Fmax  3.6 kN
Dimensions
Solution:
1.
D  33 mm
Fmin  0.90 kN
d  25 mm
r  3  mm
Determine the mean and alternating components of the fluctuating stresses.
σmb 
σab 
Tension
σxmax  σxmin
σxmax  σxmin
σat 
σab  6.150  MPa
2
σxtmax 
σmt 
σmb  10.250 MPa
2
σxtmin 
Fmax
σxtmax  48.000 MPa
h d
Fmin
σxtmin  12.000 MPa
h d
σxtmax  σxtmin
σmt  30.000 MPa
2
σxtmax  σxtmin
σat  18.000 MPa
2
Using Appendix C, determine the geometric stress concentration factors for the bending and tensile stresses.
Bending (Fig. C-10): For
D
d
r
 1.32
Ktb  A  
d
r
 0.12
Tension (Fig. C-9): For
D
d
Ktt  A  
d
r

d
b  0.27269
Ktb  1.709
r
 1.32
A  0.95880
b

d
3.
h  3  mm
See Figure P6-21 and Mathcad file P0662.
Bending
2.
σxmin  4.1 MPa
 0.12
A  1.05440
b  0.27021
b
Ktt  1.87
Calculate the notch sensitivity of the material for bending and tension using Table 6-6.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-62-2
Bending and tension:
2
Neuber constant (for S ut  55.1 ksi)
Notch sensitivity
a  0.118  in
1
q 
1
4.
5.
Bending
Kfb  1  q   Ktb  1 
Kfb  1.528
Tension
Kft  1  q   Ktt  1 
Kft  1.648
Determine what, if any, fatigue stress concentration factor should be applied to the mean stresses.
Evaluate
S y  207  MPa
Kft   σxmax  σxtmax  106.1  MPa
Kfbm  Kfb
which is less than S y so
Kftm  Kft
Calculate the mean and alternating components of the stresses increased by the appropriate fatigue stress
concentration factors.
Bending
Tension
7.
r
Calculate the fatigue stress concentration factors for bending and tension using equation 6.11b.
Yield strength SAE 1040 CR
6.
q  0.744
a
σmb  Kfb σmb
σmb  15.662 MPa
σab  Kfb σab
σab  9.397  MPa
σmt  Kftm σmt
σmt  49.427 MPa
σat  Kft  σat
σat  29.656 MPa
Find the mean and alternating von Mises stresses using equations 6.22b (with y = 0).
Mean
σ'm  σmb  σmt
σ'm  65.089 MPa
Alternating
σ'a  σab  σmb
σ'a  25.06  MPa
8. Calculate the unmodified endurance limit.
S'e  0.5 S ut
S'e  190  MPa
9. Calculate the endurance limit modification factors for a rotating, round shaft.
Load
Cload  0.70
Size
Csize  1
Surface
A  4.51
Csurf
 Sut 
 A  

 MPa 
b  0.265
(machined)
b
Csurf  0.934
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
Temperature
Ctemp  1
Reliability
Creliab  0.897
6-62-3
(R = 90%)
10. Calculate the modified endurance limit.
S e  Cload  Csize Csurf  Ctemp Creliab S'e
S e  111.48 MPa
11. Assuming a Case 3 load line, determine the factor of safety against fatigue failure.
Nf 
S e S ut
σ'a S ut  σ'm S e
Nf  2.5
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-63a-1
PROBLEM 6-63a
Statement:
For a filleted flat bar in tension similar to that shown in Appendix Figure C-9 and the data from
row a from Table P6-7, determine the alternating and mean axial stresses as modified by the
appropriate stress concentration factors in the bar.
Given:
Strength SAE 1020 CR
D  40 mm
h  10 mm
Pmin  8000 N
Widths
Thickness
Force
Solution:
1.
Determine the nominal mean and alternating components of the fluctuating stresses.
Pmax
σxm 
σxa 
σxmax  160.0  MPa
h d
Pmin
σxmin 
σxmin  40.0 MPa
h d
σxmax  σxmin
σxm  100.0  MPa
2
σxmax  σxmin
σxa  60.0 MPa
2
Using Appendix C-9, determine the geometric stress concentration factor.
For
D
d
r
2
d
Kt  A  
r
 0.2
A  1.09960
b  0.32077
b

d
3.
d  20 mm
Radius r  4  mm
Pmax  32000  N
See Table P6-7 and Mathcad file P0663a.
σxmax 
2.
S ut  469  MPa
Kt  1.843
Calculate the notch sensitivity of the material using Table 6-6.
2
Neuber constant (for S ut  68 ksi)
Notch sensitivity
q 
a  0.096  in
1
1
4.
r
Calculate the fatigue stress concentration factor using equation 6.11b.
Kf  1  q   Kt  1 
5.
q  0.805
a
Kf  1.679
Determine what, if any, fatigue stress concentration factor should be applied to the mean stress.
Yield strength SAE 1020 CR
S y  393  MPa
Evaluate
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
Kfm 
6-63a-2
S 1  Kf  σxmax
S 2  Kf  σxmax  σxmin
return Kf if S 1  S y
return
S y  Kf  σxa
σxm
if  S 1  S y  S 2  2  S y
0 otherwise
Kfm  1.679
6.
Calculate the mean and alternating components of the stresses increased by the appropriate fatigue stress
concentration factors.
σm  Kfm σxm
σm  167.9  MPa
σa  Kf  σxa
σa  100.7  MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-64a-1
PROBLEM 6-64a
Statement:
For a filleted flat bar in bending similar to that shown in Appendix Figure C-10 and the data from
row a from Table P6-7, determine the alternating and mean bending stresses as modified by the
appropriate stress concentration factors in the bar.
Given:
Strength SAE 1020 CR
D  40 mm
d  20 mm
Radius r  4  mm
h  10 mm
Mmin  80 N  m Mmax  320  N  m
Widths
Thickness
Moment
Solution:
1.
See Table P6-7 and Mathcad file P0664a.
Determine the nominal mean and alternating components of the fluctuating stresses.
c 
d
c  10 mm
2
Mmin c
σxmin 
σxm 
σxa 
h d
3
3
I  6.667  10  mm
12
Mmax c
σxmax  480  MPa
I
σxmax  σxmin
σxm  300.0  MPa
2
σxmax  σxmin
σxa  180.0  MPa
2
D
d
r
2
d
Kt  A  
r
 0.2
A  0.93232
b  0.30304
b

d
Kt  1.518
Calculate the notch sensitivity of the material using Table 6-6.
2
Neuber constant (for S ut  68 ksi)
Notch sensitivity
q 
a  0.096  in
1
1
4.
q  0.805
a
r
Calculate the fatigue stress concentration factor using equation 6.11b.
Kf  1  q   Kt  1 
5.
4
Using Appendix C-10, determine the geometric stress concentration factor.
For
3.
I 
σxmin  120  MPa
I
σxmax 
2.
S ut  469  MPa
Kf  1.417
Determine what, if any, fatigue stress concentration factor should be applied to the mean stresses.
Yield strength SAE 1020 CR
S y  393  MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-64a-2
Evaluate
Kfm 
S 1  Kf  σxmax
S 2  Kf  σxmax  σxmin
return Kf if S 1  S y
return
S y  Kf  σxa
σxm
if  S 1  S y  S 2  2  S y
0 otherwise
Kfm  0.460
6.
Calculate the mean and alternating components of the stresses increased by the appropriate fatigue stress
concentration factors.
σm  Kfm σxm
σm  137.9  MPa
σa  Kf  σxa
σa  255.1  MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-65a-1
PROBLEM 6-65a
Statement:
For a shaft, with a shoulder fillet, in tension similar to that shown in Appendix Figure C-1 and the
data from row a from Table P6-7, determine the alternating and mean axial stresses as modified by
the appropriate stress concentration factors in the shaft.
Given:
Strength SAE 1020 CR
D  40 mm
h  10 mm
Pmin  8000 N
Widths
Thickness
Force
Solution:
1.
Determine the nominal mean and alternating components of the fluctuating stresses.
4  Pmax
π d
σxmin 
σxm 
σxa 
σxmax  101.9  MPa
2
4  Pmin
π d
σxmin  25.5 MPa
2
σxmax  σxmin
σxm  63.7 MPa
2
σxmax  σxmin
σxa  38.2 MPa
2
Using Appendix C-1, determine the geometric stress concentration factor.
For
D
d
r
2
d
Kt  A  
r
 0.2
A  1.01470
b  0.30035
b

d
3.
d  20 mm
Radius r  4  mm
Pmax  32000  N
See Table P6-7 and Mathcad file P0663a.
σxmax 
2.
S ut  469  MPa
Kt  1.645
Calculate the notch sensitivity of the material using Table 6-6.
2
Neuber constant (for S ut  68 ksi)
Notch sensitivity
q 
a  0.096  in
1
1
4.
r
Calculate the fatigue stress concentration factor using equation 6.11b.
Kf  1  q   Kt  1 
5.
q  0.805
a
Kf  1.52
Determine what, if any, fatigue stress concentration factor should be applied to the mean stress.
Yield strength SAE 1020 CR
S y  393  MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
Evaluate
Kfm 
6-65a-2
S 1  Kf  σxmax
S 2  Kf  σxmax  σxmin
return Kf if S 1  S y
return
S y  Kf  σxa
σxm
if  S 1  S y  S 2  2  S y
0 otherwise
Kfm  1.520
6.
Calculate the mean and alternating components of the stresses increased by the appropriate fatigue stress
concentration factors.
σm  Kfm σxm
σm  96.7 MPa
σa  Kf  σxa
σa  58.0 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-66a-1
PROBLEM 6-66a
Statement:
For a shaft, with a shoulder fillet, in bending similar to that shown in Appendix Figure C-2 and
the data from row a from Table P6-7, determine the alternating and mean bending stresses as
modified by the appropriate stress concentration factors in the shaft.
Given:
Strength SAE 1020 CR
D  40 mm
d  20 mm
Radius r  4  mm
h  10 mm
Mmin  80 N  m Mmax  320  N  m
Widths
Thickness
Moment
Solution:
1.
See Table P6-7 and Mathcad file P0666a.
Determine the nominal mean and alternating components of the fluctuating stresses.
c 
d
c  10 mm
2
Mmin c
σxmin 
σxm 
σxa 
π d
4
3
I  7.854  10  mm
64
Mmax c
σxmax  407.437  MPa
I
σxmax  σxmin
σxm  254.6  MPa
2
σxmax  σxmin
σxa  152.8  MPa
2
D
d
r
2
d
Kt  A  
r
 0.2
A  0.90879
b  0.28598
b

d
Kt  1.44
Calculate the notch sensitivity of the material using Table 6-6.
2
Neuber constant (for S ut  68 ksi)
Notch sensitivity
q 
a  0.096  in
1
1
4.
q  0.805
a
r
Calculate the fatigue stress concentration factor using equation 6.11b.
Kf  1  q   Kt  1 
5.
4
Using Appendix C-2, determine the geometric stress concentration factor.
For
3.
I 
σxmin  101.859  MPa
I
σxmax 
2.
S ut  469  MPa
Kf  1.354
Determine what, if any, fatigue stress concentration factor should be applied to the mean stress.
Yield strength SAE 1020 CR
S y  393  MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
Evaluate
Kfm 
6-66a-2
S 1  Kf  σxmax
S 2  Kf  σxmax  σxmin
return Kf if S 1  S y
return
S y  Kf  σxa
σxm
if  S 1  S y  S 2  2  S y
0 otherwise
Kfm  0.731
6.
Calculate the mean and alternating components of the stresses increased by the appropriate fatigue stress
concentration factors.
σm  Kfm σxm
σm  186.1  MPa
σa  Kf  σxa
σa  206.9  MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-67-1
PROBLEM 6-67
Statement:
A machine part is subjected to fluctuating, simple, multiaxial stresses. The fully corrected
nonzero stress ranges are: σxmin = 50 MPa, σxmax = 200 MPa, σymin = 80 MPa, σymax = 320 MPa,
τxymin = 120 MPa, τxymax = 480 MPa. The material properties are: S e = 525 MPa and S ut = 1200
MPa. Using a Case 3 load line, calculate and compare the infinite-life safety factors given by the
Sines and von Mises Methods.
Given:
Material properties:
σxmin  50 MPa
Stresses:
Solution:
1.
S ut  1200 MPa
S e  525  MPa
σxmax  200  MPa
σymin  80 MPa
σymax  320  MPa
τxymin  120  MPa
τxymax  480  MPa
See Mathcad file P0667.
Calculate the alternating, and mean components of the given stresses.
σxa 
σxm 
σya 
σym 
τxya 
τxym 
σxmax  σxmin
σxa  75 MPa
2
σxmax  σxmin
σxm  125 MPa
2
σymax  σymin
σya  120 MPa
2
σymax  σymin
σym  200 MPa
2
τxymax  τxymin
2
τxymax  τxymin
2
τxya  180 MPa
τxym  300 MPa
(a) Sines Method
2.
Calculate the equivalent alternating and mean stresses using equations 6.21b.
2
σ'a 
2
σxa  σya  σxa σya  3  τxya
2
σ'a  329.0 MPa
σ'm  σxm  σym
3.
σ'm  325.0 MPa
The factor of safety for the Sines Method, using equation (6.18e) is
Nfs 
S e S ut
Nfs  1.11
σ'a S ut  σ'm S e
(b) von Mises Method
4.
Calculate the equivalent alternating and mean stresses using equations 6.22b.
σ'a 
2
2
σxa  σya  σxa σya  3  τxya
2
σ'a  329.0 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
σ'm 
5.
2
2
σxm  σym  σxm σym  3  τxym
σ'm  548.3 MPa
The factor of safety for the von Mises Method, using equation (6.18e) is
Nfvm 
6.
2
6-67-2
S e S ut
σ'a S ut  σ'm S e
Nfvm  0.92
This example shows that the von Mises method is more conservative when the endurance limit is modified by
such factors as surface finish and when there is a high mean shear stress.
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-68-1
PROBLEM 6-68
Statement:
A cylindrical tank with hemispherical ends has been built. It was made from hot rolled steel that
has S ut = 380 MPa. The tank outside diameter is 300 mm with 20 mm wall thickness. The
pressure may fluctuate from 0 to an unknown maximum. For an infinite-life fatigue safety factor
of 4 with 99.99% reliability, what is the maximum pressure to which the tank may be subjected?
Given:
Ultimate strength
S ut  380  MPa
Tank dimensions
Reliability & FS
OD  300  mm
R  0.9999
Solution:
1.
t  20 mm
Nf  4
See Mathcad file P0668.
Determine the maximum principal stresses as functions of the unkown pressure, which occur at the inside wall,
for this thick-wall cylinder:
Outside radius
ro  0.5 OD
ro  150 mm
Inside radius
ri  ro  t
ri  130 mm
Tangential
σt( p ) 
Radial
2

ro 


σr( p ) 
1
2
2
2
ri 
ro  ri 
2

 1  ro 
2
2
2
ri 
ro  ri 
2
ri  p
2
ri  p
2
ri  p
σa( p ) 
Axial
2
ro  ri
2
These are principal stresses so, using equation 5.7a:
σ'( p ) 
2
2
2
σt( p )  σr( p )  σa( p )  σt( p )  σa( p )  σa( p )  σr( p )  σt( p )  σr( p )
σ'min  0  MPa
2.
Determine the alternating, and mean von Mises effective stress using equations 6.1.
σ'a( p ) 
σ'( p )  σ'min
σ'm( p ) 
2
σ'( p )  σ'min
S'e  0.5 S ut
3.
Calculate the unmodified endurance limit.
4.
Calculate the endurance limit modification factors for axial loading.
Load
Cload  0.7
(axial loading)
Size
Csize  1
(axial loading)
Surface
A  57.7
Csurf
Temperature
2
b  0.718
 Sut 
 A  

 MPa 
S'e  190  MPa
(hot rolled)
b
Csurf  0.811
Ctemp  1
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
Reliability
5.
Creliab  0.702
(R = 99.99%)
Calculate the modified endurance limit.
S e  Cload  Csize Csurf  Ctemp Creliab S'e
6.
6-68-2
S e  75.7 MPa
Assuming a Case 3 load line, solve equation (6.18e) for p max
Guess
p  1  MPa
Given
Nf =
S e S ut
σ'a( p )  S ut  σ'm( p )  S e
p max  Find ( p )
7.
Nf  4.00
p max  4.54 MPa
The principal stresses at p max are, respectively:
σ1  σt p max 
σ2  σa p max 
σ3  σr p max 
σ1  31.9 MPa
σ2  13.7 MPa
σ3  4.5 MPa
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-69-1
PROBLEM 6-69
Statement:
A rotating shaft has been designed and fabricated from SAE 1040 HR steel. It is made from
tubing that has an outside diameter of 60 mm and a wall thickness of 5 mm. Strain gage
measurements indicate that there is a fully reversed axial stress of 68 MPa and a torsional stress
that fluctuates from 12 MPa to 52 MPa in phase with the axial stress at the critical point on the
shaft. Determine the infinite-life fatigue safety factor for the shaft for a reliability of 99%.
Given:
Strength SAE 1040 HR
S ut  524  MPa Alternating bending stress σxa  68 MPa
Torsional stress
τxymax  52 MPa
τxymin  12 MPa
Shaft dimensions
OD  60 mm
t  5  mm
See Mathcad file P0669.
Solution:
1. Calculate the alternating, and mean components of the given stresses.
2.
σxm  0  MPa
τxymax  τxymin
τxya 
σxa  68 MPa
τxym 
τxym  32 MPa
τxya  20 MPa
2
τxymax  τxymin
2
Find the mean and alternating von Mises stresses using equations 6.22b (with y = 0).
2
Mean
σ'm 
3  τxym
σ'm  55.426 MPa
Alternating
σ'a 
σxa  3  τxya
2
2
σ'a  76.315 MPa
3. Calculate the unmodified endurance limit.
S'e  0.5 S ut
S'e  262  MPa
4. Calculate the endurance limit modification factors for a rotating, round shaft.
Load
Cload  1
Size
OD 
Csize  1.189  

 mm 
Surface
A  57.7
Csurf
 Sut 
 A  

 MPa 
Temperature
Ctemp  1
Reliability
Creliab  0.814
(combined bending and torsion)
 0.097
Csize  0.799
b  0.718
(hot rolled)
b
Csurf  0.644
(R = 99%)
5. Calculate the modified endurance limit.
S e  Cload  Csize Csurf  Ctemp Creliab S'e
S e  110  MPa
6. Assuming a Case 3 load line, determine the infinite-life fatigue safety factor.
Nf 
S e S ut
σ'a S ut  σ'm S e
Nf  1.25
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-70a-1
PROBLEM 6-70a
Statement:
For the data in row a in Table P6-8, find the safety factor for each of the four loading cases based
on the Modified-Goodman diagram if S e = 100, S y = 150, and S ut = 200.
Given:
S e  100
S y  150
σ'm  50
σ'a  30
Solution:
1.
S ut  200
See Table 6-8 and Mathcad file P0670a.
Calculate the coordinates of point D in Figure 6-46 using equations (6.16).
Dm 
S ut  S y  S e
Dm  100
S ut  S e
Da  S y  Dm
1.
Da  50
Use equations (6.18) and (6.16) to calculate the required quantities.
Case 1:
Nf1 

Sy
σ'a 
1 
 if σ'a  Da

S ut 
σ'a 
1 
 otherwise
σ'm 
Se 
σ'm 
Nf1  2.40
Sy
Case 2:
Nf2 
Se

1 
σ'm 
 if σ'm  Dm
Nf2  2.50
σ'a 
S ut 
S y  σ'm
otherwise
σ'a
Case 3:
Nf3 
S e S ut
σ'a S ut  σ'm S e
if
Se
σ'a
σ'm
S y  σ'a
 Dm
Nf3  1.82
S ut
otherwise
σ'm
Case 4:

Se
S ut  S e  S e σ'a  S ut σ'm
2
σ'mSut 
2
S e  S ut
σ'aSut  S e 
OZ 
2
Se
S ut
2
 σ'mSut
2
σ'a  σ'm
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-70a-2
 σ'm  σ'mSut 2  σ'a  σ'aSut2
ZSut 
Se
if σ'm 
σ'a
σ'm

σ'mSy 
σ'm  σ'mSut2   σ'a  σ'aSut 2

Se
S ut
otherwise
S y  σ'a  σ'm
2
σ'aSy  σ'mSy  σ'a  σ'm
 σ'm  σ'mSy2   σ'a  σ'aSy 2
ZSy 
Sy
if σ'm 
1

Nf4 
σ'm  σ'mSy 2  σ'a  σ'aSy2
OZ  ZSut
OZ
OZ  ZSy
OZ
if σ'mSut  Dm
σ'a
σ'm
otherwise
Nf4  1.69
otherwise
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-70h-1
PROBLEM 6-70h
Statement:
For the data in row h in Table P6-8, find the safety factor for each of the four loading cases based
on the Modified-Goodman diagram if S e = 100, S y = 150, and S ut = 200.
Given:
S e  100
S y  150
σ'm  80
σ'a  80
Solution:
1.
S ut  200
See Table 6-8 and Mathcad file P0670h.
Calculate the coordinates of point D in Figure 6-46 using equations (6.16).
Dm 
S ut  S y  S e
Dm  100
S ut  S e
Da  S y  Dm
1.
Da  50
Use equations (6.18) and (6.16) to calculate the required quantities.
Case 1:
Nf1 

Sy
σ'a 
1 
 if σ'a  Da

S ut 
σ'a 
1 
 otherwise
σ'm 
Se 
σ'm 
Nf1  0.50
Sy
Case 2:
Nf2 
Se

1 
σ'm 
 if σ'm  Dm
Nf2  0.75
σ'a 
S ut 
S y  σ'm
otherwise
σ'a
Case 3:
Nf3 
S e S ut
σ'a S ut  σ'm S e
if
Se
σ'a
σ'm
S y  σ'a
 Dm
Nf3  0.83
S ut
otherwise
σ'm
Case 4:

Se
S ut  S e  S e σ'a  S ut σ'm
2
σ'mSut 
2
S e  S ut
σ'aSut  S e 
OZ 
2
Se
S ut
2
 σ'mSut
2
σ'a  σ'm
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-70h-2
 σ'm  σ'mSut 2  σ'a  σ'aSut2
ZSut 
Se
if σ'm 
σ'a
σ'm

σ'mSy 
σ'm  σ'mSut2   σ'a  σ'aSut 2

Se
S ut
otherwise
S y  σ'a  σ'm
2
σ'aSy  σ'mSy  σ'a  σ'm
 σ'm  σ'mSy2   σ'a  σ'aSy 2
ZSy 
Sy
if σ'm 
1

Nf4 
σ'm  σ'mSy 2  σ'a  σ'aSy2
OZ  ZSut
OZ
OZ  ZSy
OZ
if σ'mSut  Dm
σ'a
σ'm
otherwise
Nf4  0.84
otherwise
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
6-71-1
PROBLEM 6-71
Statement:
A rotating shaft with a shoulder fillet, in torsion similar to that shown in Appendix Figure C-3 is
made from SAE 1020 CR steel and has dimensions D = 40 mm, d = 20 mm, and r = 4 mm. The
shaft is ground and is subjected to a fully reversed torque of +/- 80 N-m. Determine the infinite
life safety factor for the shaft for a reliability of 99.9%.
Given:
Strength SAE 1020 CR
D  40 mm
Ta  80 N  m
Dimensions
Torque
Solution:
1.
S ut  469  MPa
d  20 mm
Tm  0  N  m
r  4  mm
See Mathcad file P0671.
Determine the nominal mean and alternating components of the stresses.
τxym  0  MPa
τxya 
16 Ta
π d
2.
τxya  50.9 MPa
3
Using Appendix C-3, determine the geometric stress concentration factor.
For
D
Kt  A  
r
2
d
r
d
 0.2
A  0.86331
b

d
3.
Kt  1.268
Calculate the notch sensitivity of the material using Table 6-6 adding 20 ksi to S ut because of the torsional load
Neuber constant (for S ut  20 ksi  88 ksi)
q 
Notch sensitivity
1
1
4.
6.
q  0.846
a
r
Kf  1.226
Calculate the mean and alternating components of the stresses increased by the appropriate fatigue stress
concentration factors.
τm  τxym
τm  0  MPa
τa  Kf  τxya
τa  62.46  MPa
Calculate the von Mises normal stress.
von Mises stress
7.
2
a  0.072  in
Calculate the fatigue stress concentration factor using equation 6.11b.
Kf  1  q   Kt  1 
5.
b  0.23865
σ'a 
3  τa
σ'a  108.19 MPa
Calculate the endurance limit modification factors for a rotating, solid, round steel shaft.
Load
Cload  1
(pure torsion)
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P0671.xmcd
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
Size
Csize  1.189  


 mm 
Surface
A  1.58
Csurf
d
6-71-2
 0.097
Csize  0.889
b  0.085
 Sut 
 A  

 MPa 
Temperature
Ctemp  1
Reliability
Creliab  0.753
(ground)
b
Csurf  0.937
(R = 99.9%)
S'e  0.5 S ut
Uncorrected endurance strength
S'e  234.5  MPa
8.
Calculate the endurance limit.
S e  Cload  Csize Csurf  Ctemp Creliab S'e
9.
S e  147.1  MPa
Using equation 6.14, calculate the infinite-life factor of safety.
Nf 
Se
σ'a
Nf  1.4
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
7-1-1
PROBLEM 7-1
Statement:
Two 3 x 5 cm blocks of steel with machined finish Ra = 0.6 mm are rubbed together with a normal
force of 400 N. Estimate the true area of contact between them if their S y = 400 MPa.
Given:
Length of block
Width of block
L  5  cm
w  3  cm
Normal force
Yield strength
F  400  N
S y  400  MPa
Assumptions: The compressive yield strength is the same as the tensile yield strength. Then, S yc  S y.
Solution:
1.
See Mathcad file P0701.
Using equation 7.1, the true area of contact is estimated as
Ar 
3
F
Ar  3.33333  10
3  S yc
2.
The apparent area of contact is
3.
The ratio of apparent area to true area is
Aa  w L
2
cm
2
Aa  15 cm
Aa
Ar
 4500
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
7-2-1
PROBLEM 7-2
Statement:
Given:
Estimate the dry coefficient of friction between the two pieces in Problem 7-1 if their S ut = 600
MPa.
Length of block
L  5  cm
Yield strength
S y  400  MPa
Width of block
w  3  cm
Ultimate strength
S ut  600  MPa
Normal force
F  400  N
Assumptions: The compressive yield strength is the same as the tensile yield strength. Then, S yc  S y.
Solution:
See Mathcad file P0702.
Using equations 7.3 and 7.4, the coefficient of friction is estimated as
S us  0.80 S ut
μ 
S us
3  S yc
S us  480 MPa
μ  0.40
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
7-3-1
PROBLEM 7-3
Statement:
For the bicycle pedal-arm assembly in Figure P7-1 assume a rider-applied force that ranges from
0 to 400 N at the pedal each cycle. Determine the maximum contact stresses at one sprocket
tooth-chain roller interface. Assume that the one tooth takes all the applied torque, that the
chain roller is 8-mm dia, the sprocket has a nominal (pitch) dia of 100 mm, and that the sprocket
tooth is essentially flat at the point of contact. The roller and sprocket are made of SAE X1340
steel, induction hardened to HRC 45-58. The roller and sprocket contact over a length of 8 mm.
Assuming rolling plus 9% sliding, estimate the number of cycles to failure for this particular
tooth-roller combination.
Given:
Roller radius
R1  4  mm
Sprocket radius
R2  ∞ mm
Sprocket width
Pedal force
w  8  mm
Frider  400  N
The parts are steel. Therefore:
E  207  GPa
ν  0.28
Sprocket pitch dia
d p  100  mm
Pedal arm length len  170  mm
Assumptions: The coefficient of friction is μ  0.33
Solution:
1.
2.
See Figure P7-1 and Mathcad file P0703.
Determine the maximum contact force.
Torque on sprocket
Tmax  Frider len
Contact force
Fcmax 
2  Tmax
Tmax  68 N  m
Fcmax  1.36 kN
dp
Find the material constants from equation 7.9a.
2
Material constants
m1 
1ν
6
m1  4.452  10
E
1
MPa
m2  m1
Geometry constant
B 
1
2
 
1
 R1

1


1
B  125 m
R2 
1
Contact patch
half-width
3.
4.
a 
 2 m1  m2 Fcmax 

 

B
w 
π
2
a  0.0878 mm
The average and maximum contact pressure can now be found from equations 7.14b and c.
Fcmax
Average pressure
p avg 
Maximum pressure
p max 
Tangential
pressure
fmax  μ  p max
2 a w
2  Fcmax
π a  w
p avg  968.1 MPa
p max  1233 MPa
fmax  406.8 MPa
With m = 0.33, the principal stresses in the contact zone will be maximal on the surface (z = 0) at x = 0.3a from th
centerline as shown in Figures 7-20 and 7-22. The applied stress components are found from equation 7.23a for
the normal force and equation 7.23b for the tangential force.
For
x  0.3 a
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
σxn  p max  1 
x
2
a
σxt  2  fmax 
σxn  1176 MPa
2
x
σxt  244.05 MPa
a
σzn  p max  1 
x
2
a
σzn  1176 MPa
2
σzt  0  MPa
τxzt  fmax  1 
τxzn  0  MPa
x
2
a
5.
6.
τxzt  388.0 MPa
2
Equations 7.24a and 7.24b can now be solved for the total applied stresses along the x, y, and z axes.
σx  σxn  σxt
σx  1420 MPa
σz  σzn  σzt
σz  1176 MPa
τxz  τxzn  τxzt
τxz  388.019 MPa
Assuming the rollers are short, we expect a plane stress condition to exist. The stress in the third dimension is
then:
σy  0  MPa
7.
7-3-2
also,
τxy  0  MPa
τyz  0  MPa
Unlike the pure-rolling case, these stresses are not principal because of the applied shear stress. The principal
stresses are found from equation 4.4 using a cubic root finding solution.
C2 
σx  σy  σz
C2  2.596  10
MPa
τxz 
τyz 
 σx τxy 
 σx
 σy






MPa MPa 
MPa MPa 
MPa MPa 



C1 


 τxy
 τxz σz 
 τyz σz 
σy 
 MPa MPa 
 MPa MPa 
 MPa MPa 






C1  1.519  10
 σx τxy τxz 
 MPa MPa MPa 
 τxy
σy
τyz 
C0  

 MPa MPa MPa 
 τxz τyz σz 
 MPa MPa MPa 


C0  0
3
3
6
2
f ( σ )  σ  C2 σ  C1 σ  C0
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MACHINE DESIGN - An Integrated Approach, 4th Ed.
 C0 


C1 

v 