ME 360 – H06
Name _________________________
1) Norton text problem 6-61. Use Case 3) Goodman failure.
See PDF solution
May skip Step 5 in PDF solution
Substitute for Step 11 in PDF solution
SN = Se
Nalt = SN / σalt’ = 1.791
Nmean = Sut / σmean’ = 6.642
NGOODMAN = Nalt Nmean / (Nalt + Nmean) = 1.41
ME 360 – H06
Name _________________________
2) Check factor of safety for fatigue failure of Norton text problem 5-73 after five years of
continuous operation at crank speed 15 rpm. Use 1/2x3/4 inch 6061-T6 aluminum bar stock
available from Online Metals. Do not include stress concentrations for connections at A and B.
Use Falt=1850 lbf and Fmean=1650 lbf. Use Case 3) Goodman failure.
 15 rev  60 min  24 hr  365 days 

 = 39.42x106 rev
5 years


min
hr
day
year





must use S-N diagram
0.5 x 0.75 inch rectangular bar, 6061-T6 extruded aluminum
Sm = 0.75 Sut = 33.75 ksi
Sf’ = 0.4 Sut = 18.0 ksi
axial loading
CLOAD = 0.70
CSIZE = 1.0
axial loading
Eq6.5c Norton
Eq 6.6 Norton
Eq 6.7a Norton
axial loading
footnote Norton
CSURF =A Sut b = 0.9846 A = 2.7 b = -0.265
assume extruded similar to cold-rolled
CTEMP = 1.0
CRELI = 0.814
assume room temperature
assume R = 99%
Table 6-3 and Eq 6.7e Norton
Eq 6.7f Norton
Table 6-4 Norton
Sf = (0.70) (1.0) (0.9846) (1.0) (0.814) (18.0 ksi) = 10.10 ksi
A = (0.5 in) (0.75 in) = 0.375 in2
σalt = Falt / A = 4.933 ksi
Online Metals
Eq6.9 Norton
at N = 5x108 cycles
Sf = CLOAD CSIZE CSURF CTEMP CRELI Sf’
Sut = 45.0 ksi
σmean = Fmean / A = 4.40 ksi
ME 360 – H06
Name _________________________
logS
Sm =
33.75 ksi
Sut = 45 ksi
SN
Sf =
10.10 ksi
logN
103
N=39.42x106
log S N − log S m
log N − log 10 3
=
8
3
log S f − log S m
log 5x10 − log 10
(
)
5x108
0.8064 =
log S N − 1.5283
− 0.52395
SN = 12.76 ksi
Nalt = SN / σalt = 2.587
Nmean = Sut / σmean = 10.227
NGOODMAN = Nalt Nmean / (Nalt + Nmean ) = 2.06
3) Sketch and label applied load versus time for problem 2) above and compare to the load for
Norton text problem 5-73.
F
4 sec
Falt = 1850 lbf
Fmean = 1650 lbf
time
ME 360 – H06
Name _________________________
4) Briefly describe the differences between 1040 and 4140 steels that have been quenched and
tempered at 400° F in terms of yield strength, ultimate strength, chemical composition and cost.
Remember to cite your sources. Create a table and provide typical numerical values for strength,
compostion and cost. Remember to cite your sources.
AISI 1040
quenched and
tempered 400°F
AISI 4140
quenched and
tempered 400°F
Source
SY
86 ksi
238 ksi
Norton A-9
SUT
113 ksi
257 ksi
SUT / SY
1.31
1.08
SY 4140/1040
2.76
SUT 4140/1040
2.27
Iron
Carbon
Manganese
mostly
0.37 - 0.44 %
0.60 - 0.90 %
Chromium
Silicon
Molybdenum
mostly
0.38 - 0.43 %
0.75 - 1.00 %
Azom.com
0.80 - 1.10 %
0.15 - 0.30 %
0.15 - 0.25 %
Sulphur
Phosphorus
< 0.05 %
< 0.04 %
< 0.04 %
< 0.035 %
1 inch DIA
12 inch long
$9.91
(1045 annealed)
$13.55
(annealed)
McMaster-Carr
1) AISI (SAE) 4140 is over twice as strong as 1040 in both yield and ultimate strength. AISI
1040 is more ductile with ultimate failure 31 % higher than yield while 4140 is less
ductile with ultimate failure only 8 % higher than yield.
2) Both 1040 and 4140 have approximately 0.4 % carbon and 0.8 % manganese. Additionally
AISI 4140 has approximately 1.0 % chromium with traces of molybdenum and silicon.
ME 360 – H06
Name _________________________
3) AISI 4140 is about 37% more expensive that 1040. Both are typically sold in annealed
condition.
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.
© 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,
P0661.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.
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)
© 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,
P0661.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.
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
© 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,
P0661.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.