1. No category

ME 452 - Machine Design II
Name of Student:_____________________________
Spring Semester 2015
Lecture Division Number:______________________
Homework No. 4 (50 points). Due at the beginning of lecture on Friday, February 13th.
Solve the following problems from Chapter 6, Shigley’s Mechanical Engineering Design, Ninth
Edition, R.G. Budynas and J.K. Nisbett.
1. (15 Points). Problem 6-15, page 349.
2. (15 Points). Problem 6-16, page 349.
3. (20 Points). Problem 6-17, page 349.
1
Solution to Homework Set 4.
1. Problem 6-15. (15 Points). The fully corrected endurance strength of the solid steel round bar from
Eq. (6-18), see page 287, can be written as
Se  ka kb kc kd ke k f Se 
(1)
The uncorrected endurance strength of the bar can be written from Eq. (6-8), see page 282, as
Se   0.5 Sut
(2a)
The ultimate tensile strength of AISI 1095 HR steel from Table A-20, see page 1040, is
Sut  120 kpsi
(2b)
Substituting Eq. (2b) into Eq. (2a), the uncorrected endurance strength of the solid steel round bar is
Se   0.5  120  60 kpsi
(3)
The Marine factors. The surface modification factor can be written from Eq. (6-19), see page 287, as
ka  a Sutb
(4)
The coefficient and the exponent from Table 6-2, see page 288, respectively, are
a  2.70 kpsi
and
b   0.265
(5)
Substituting Eqs. (2b) and (5) into Eq. (4), the surface modification factor is
ka  2.70 1200.265  0.759
(6)
The size modification factor. Since the round bar (diameter d  1.8 inches) is not rotating then the
equivalent diameter can be written from Eq. (6-24), see page 289, as
d e  0.370 d  0.370 1.8  0.666 inches
(7)
Therefore, the size modification factor can be written from Eq. (6-20), see page 288, as
kb  0.879 d e 0.107
(8)
Substituting Eq. (7) into Eq. (8), the size modification factor is
kb  0.879  0.6660.107  0.918
(9)
The load modification factor for fully reversed bending, see Figure 6.23f, page 301, or repeated bending,
see Figure 6.23e, page 301, from Eq. (6-26), see page 290, is
kc  1
(10a)
The temperature modification factor, see pages 290-292, is
kd  1
2
(10b)
The reliability modification factor, see page 292, is
ke  1
(10c)
The miscellaneous effects modification factor, see page 293, is
kf 1
(11)
Substituting Eqs. (3), (6), (9), (10), and (11) into Eq. (1), the fully corrected endurance strength is
Se  0.759  0.918 1 60  41.806 kpsi
(12)
Stress Analysis. The fatigue stress concentration factor can be written from Eq. (6-32), see page 295, as
K f  1  q( K t  1)
(13)
First, determine the theoretical stress concentration factor. The known geometric data for the bar is
D  2 in,
d  1.8 in
and
r  0.1 in
(14a)
Therefore, the geometric ratios are
r 0.1

 0.056
d 1.8
D
2

 1.111
d 1.8
and
(14b)
The theoretical stress concentration factor from Table A-15-14, see page 1030, is
K t  2.1
(14c)
The notch sensitivity can be obtained from either Fig. 6-20, or from the curve-fit, see Eqs. (6-34) and (635a), page 296. Using the equations, the Neuber constant from Eq. (6-35a), page 296, is
a  0.246  (3.08  103 ) Sut  (1.51 105 ) Sut 2  (2.67 108 ) Sut 3 in
(15a)
where the ultimate tensile strength in Eq. (15a) must be in kpsi. Substituting Eq. (2) into Eq. (15a), the
Neuber constant is
a  0.246  (3.08 103 )120  (1.51105 )1202  (2.67  108 )1203  0.0477 in
(15b)
The notch sensitivity can be written from Eq. (6-34), page 296, as
q
1
a
1
r
(16a)
Substituting Eqs. (14a) and (15b) into Eq. (16a), the notch sensitivity is
q
1
 0.869
0.0477
1
0.1
3
(16b)
Substituting Eqs. (14c) and (16b) into Eq. (13), the fatigue stress concentration factor is
K f  1  0.869(2.1  1)  1.956
(17)
The maximum and minimum normal stresses due to the maximum and the minimum bending
moments can be written, respectively, as
 max  
M max c
I
 min 
and
M min c
I
(18)
The maximum and the minimum bending moments, respectively, are
M max   25, 000 lb.in
and
M min  0
(19a)
The area moment of inertia of a solid round bar is
I
d4
64

 x 1.84
64
 0.515 in 4
(19b)
Substituting Eqs. (19a), (19b) into Eq. (18), the maximum normal stress and the minimum normal stress
due to the maximum and minimum bending moments, respectively, are
 max  
25, 000  (1.8 / 2)
  43.689 kpsi
0.515
and
 min  0
(20a)
(20b)
This is commonly referred to as the repeated stress problem, see page 301. The amplitude ratio A = 1
and the stress ratio R = 0. The nominal values of the mean and alternating components of the normal
stress from Eq. (6-36), see page 301, respectively, are
 m ,nom 
and
 a ,nom 
 max   min
2
 max   min
2

 43.689  0
  21.845 kpsi
2
(21a)

 43.689  0
 21.845 kpsi
2
(21b)
The mean and alternating components of the normal stress, including the fatigue stress concentration
factor, can be written as
    min 
 m  K f  m,nom  K f  max
(22a)

2


and
   min
 a  K f  a ,nom  K f max
(22b)
2
Substituting Eqs. (17) and (20) or (21) into Eqs. (22), the mean and alternating components of the stress
components of the normal stress, respectively, are
4
  43.689  0 
   42.728 kpsi
2


(23a)
 43.689  0
 42.728 kpsi
2
(23b)
 m  1.956 
and
 a  1.956
For purposes of comparison two criteria of fatigue failure for infinite life will be presented: (i) the
modified-Goodman line; and (ii) the Gerber parabola. (i) The modified-Goodman line, see Eq. (6-46),
page 306, can be written as
 
1
(24)
 a m
Nf
Se Sut
Substituting Eqs. (2), (12), (23a), and (23b) into Eq. (24), the factor of safety guarding against fatigue
failure can be written as
1
42.728 42.728
(25a)


N f 41.806
120
Rearranging this equation, the factor of safety guarding against fatigue failure is
N f  0.73
(25b)
Since the fatigue factor of safety is less than one then the solid steel round bar has a finite life.
(ii) The Gerber parabola, see Eq. (6-47), page 306, or Table 6-7, page 307, can be written as
1/ 2

2  m Se 2  
1  a Sut 2 
N f  ( )( ) 1  1  (
)  
2 Se  m 
 a Sut  



(26a)
Substituting Eqs. (2), (12), (23a) and (23b) into Eq. (26a), the factor of safety guarding against fatigue
failure can be written as
1/ 2

1 42.728 120 2 
2 x 42.728 x 41.806 2  
Nf  (
)(
) 1  1  (
)  
2 41.806 42.728 
42.728 x120

 

(26b)
or as
Nf 
1/ 2

1
1202
41.806 2  

(
) 1  1  (
)  
2 41.806 x 42.728 
60

 
(26c)
Therefore, the factor of safety guarding against fatigue failure is
N f  0.89
(26d)
Note that the answer given by Eq. (26d) is more accurate but less conservative than the answer given by
Eq. (25b). Since the fatigue factor of safety given by both Eqs. (25b) and (26d) is less than one then the
solid steel round bar is predicted to have a finite life. Therefore, the number of cycles to failure must be
determined.
5
The number of cycles to failure can be written from Eq. (6-16) see page 285, as
 K f  rev 
N 

 a 
1/ b
 K fa 


 a 
1/ b
(27a)
Alternatively, if the stress concentration effects are ignored then the number of cycles to failure can be
written as
 
N  a 
 a 
1/b
(27b)
The constant can be written from Eq. (6-14), see page 285, as
( f Sut ) 2
a
Se
(28a)
The fatigue strength fraction from Fig. 6-18, see page 285, is
f  0.82
(28b)
Substituting Eqs. (2), (12), and (28b) into Eq. (28a), the coefficient is
a
(0.82  120) 2
 231.607 kpsi
41.806
(28c)
The exponent can be written from Eq. (6-15), see page 285, as
 f Sut 
1
b   log 

3
 Se 
(29a)
Substituting Eqs. (2), (12), and (28b) into Eq. (29a), the exponent is
1
 0.82  120 
b   log 
   0.124
3
 41.806 
(29b)
Substituting Eqs. (17), (23b), (28c), and (29b) into Eq. (27a), the number of cycles to failure is
1
 1.956  42.728   0.124
N 
 3728.733 cycles  3.729 103 cycles

 231.607 
(30a)
Alternatively, substituting Eqs. (23b), (28c), and (29b) into Eq. (27b), the number of cycles to failure is
1
 42.728   0.124
N 
 831108 cycles  8.311105 cycles

231.607


(30b)
The factor of safety guarding against yielding. The stress concentration factors for static yielding of a
ductile material will not be included here, see Section 3.13, page 110. The factor of safety guarding
against yielding can be written from the Langer line, see Eq. (6-49), page 306, as
6
Ny 
Sy
(31)
 a ,nom   m ,nom
The yield strength of AISI 1095 HR steel, from Table A-20, see page 1040, is
S y  66 kpsi
(32)
Substituting Eqs. (21) and (32) into Eq. (31), the factor of safety guarding against yielding is
Ny 
66
 1.51
21.845  21.845
(33)
Since the factor of safety is greater than one then yielding is not predicted to have occurred.
Conservative Approach. The factor of safety guarding against yielding can also be written as
Ny 
Sy
(34a)
 max
Substituting Eqs. (20a) and (32) into Eq. (34a), the factor of safety guarding against yielding is
Ny 
66
 1.51
43.689
(34b)
Note that, in this case, the answer is the same as Eq. (33).
2. Problem 6-16. (15 Points). The corrected endurance strength of the shaft from Eq. (6-18), see page
287, can be written as
Se  ka kb kc kd ke k f Se 
(1)
The uncorrected endurance strength of the shaft can be written from Eq. (6-8), see page 282, as
Se   0.5Sut
(2a)
The ultimate tensile strength of AISI CD 1020 steel, from Table A-20, see page 1040, is
Sut  470 MPa 
470
kpsi  68.2 kpsi
6.89
(2b)
Substituting Eq. (2b) into Eq. (2a), the uncorrected endurance strength of the shaft is
Se   0.5  470  235 MPa
(2c)
The surface modification factor can be written from Eq. (6-19), see page 287, as
ka  a Sutb
(3a)
The coefficient and the exponent for cold drawn surface finish from Table 6-2, see page 288, are
a  4.51 MPa
and
7
b   0.265
(3b)
Substituting Eq. (3b) into Eq. (3a), the surface modification factor is
ka  4.51 4700.265  0.883
(3c)
The size modification factor from Eq. (6-20), see page 288, is
kb  1.24d 0.107  1.24  350.107  0.848
(4)
The load modification factor from Eq. (6-26), see page 290, is
kc  1
(5)
The temperature modification factor, see pages 290-292, is
kd  1
(6a)
The reliability modification factor, see page 292, is
ke  1
(6b)
The miscellaneous effects modification factor, see page 293, is
kf 1
(6c)
Substituting Eqs. (2c), (3c)-(6c) into Eq. (1), the corrected endurance strength of the shaft is
Se  0.883  0.848 1 235  175.964 MPa
(7)
Stress Analysis. The fatigue stress concentration factor can be written from Eq. (6-32), see page 295, as
K f  1  q( K t  1)
(8)
First, determine the theoretical stress concentration factor. The known geometric data for the shaft is
D  50 mm,
d  35 mm
and
r  3 mm
(9a)
The geometric ratios are
r
3

 0.086
d 35
and
D 50

 1.43
d 35
(9b)
Therefore, the theoretical stress concentration factor from Figure A-15-9, see page 1028, is
K t  1.7
(10)
The notch sensitivity can be obtained from either Fig. 6-20, or from the curve-fit, see Eqs. (6-34) and (635a), page 296. The equations will be used here. Therefore, the notch sensitivity can be written from Eq.
(6-34), page 296, as
1
q
(11)
a
1
r
8
The Neuber constant can be written from Eq. (6-35a), page 296, is
a  0.246  (3.08  103 ) Sut  (1.51 105 ) Sut 2  (2.67 108 ) Sut 3 in
(12a)
Note that the units in Eq. (12a) are in because the ultimate tensile strength is in kpsi. Substituting Eq.
(2b) into Eq. (12a), the Neuber constant is
a  0.246  (3.08 103 )68.2  (1.51 105 )68.22  (2.67 108 )68.23  0.09771 in
(12b)
Then in SI units (see Example 6-6, page 296), the Neuber constant is
a  0.09771 in  0.492 mm
(12c)
Substituting Eq. (12c) into Eq. (11), the notch sensitivity is
q
1
 0.779
0.492
1
3
(13)
Then substituting Eqs. (10) and (13) into Eq. (8), the fatigue stress concentration factor is
K f  1  0.779  (1.7  1)  1.545
(14)
The free-body diagram of the shaft is shown in Figure 1.
Figure 1. The free body diagram of the shaft.
The sum of the reaction forces and the sum of the moments about the right-hand bearing can be written,
respectively, as
R1  R2  6 kN
and
500 R1  175 x 6  0
(15a)
Therefore, the reaction forces at the left bearing and the right bearing, respectively, are
R1 = 2.1 kN
and
R2 = 3.9 kN
(15b)
The critical section of the shaft will be at the shoulder fillet between the 35 mm and 50 mm
diameters where the bending moment is large, the diameter is smaller, and stress concentration exists.
For the rotating shaft, the loading is completely reversed bending.
The factor of safety guarding against yielding can be written as
9
Ny 
Sy
 rev
(16a)
The yield strength of AISI 1020 CD steel, from Table A-20, see page 1040, is
S y  390 MPa
(16b)
The normal stress due to the completely reversed bending moment can be written as
 rev 
Mc
I
(17a)
The completely reversed bending moment on the critical element at the critical section is
M  200  2.1  420 Nm
(17b)
Substituting Eq. (17b) and the known geometry into Eq. (17a), the normal stress due to the completely
reversed bending moment is
 rev 
420  (35 / 2)
 0.09978 kN/mm 2  99.78 MPa
4
( / 64)  35
(18)
Substituting Eqs. (16b) and (18) into Eq. (16a), the factor of safety guarding against yielding is
Ny 
390
 3.91
99.78
(19)
Note. The completely reversed bending stress is much less than the yield strength of the shaft material,
therefore, yielding is not predicted to have occurred.
Since this problem is a zero mean stress problem then the factor of safety guarding against fatigue
failure can be written as
S
Se
Nf  e 
(20a)
 a K f  rev
Substituting Eqs. (7), (14), and (17) into Eq. (20a), the fatigue factor of safety is
Nf 
175.964
 1.14
1.545  99.78
(20b)
Since the factor of safety guarding against fatigue failure is greater than one then the shaft has an infinite
life, that is, infinite life is predicted.
3. Problem 6-17. (20 Points). The corrected endurance strength of the shaft from Eq. (6-18), see page
287, can be written as
Se  ka kb kc kd ke k f Se 
(1)
The ultimate tensile strength of AISI 1040 CD steel, from Table A-20, see page 1040, is
Sut  85 kpsi
10
(2)
The uncorrected endurance strength of the shaft can be written from Eq. (6-8), see page 282, as
Se   0.5Sut  0.5  85  42.5 kpsi
(3)
The surface modification factor from Eq. (6-19), see page 287, can be written as
ka  aSutb
(4a)
The coefficient and the exponent from from Table 6-2, see page 288, are
a  2.7 kpsi
and
b   0.265
(4b)
Substituting Eqs. (4b) into Eq. (4a), the surface modification factor is
ka  2.7  850.265  0.832
(5)
The size modification factor from Eq. (6-20), see page 288, is
kb  0.879d 0.107  0.879 1.6250.107  0.835
(6)
The load modification factor from Eq. (6-26), see page 290, is
kc  1
(7a)
The temperature modification factor, see pages 290-292, is
kd  1
(7b)
The reliability modification factor, see page 292, is
ke  1
(7c)
The miscellaneous effects modification factor, see page 293, is
kf 1
(7d)
Substituting Eqs. (3) and (5)-(7d) into Eq. (1), the corrected endurance strength of the shaft is
Se  0.832  0.835 1 42.5  29.526 kpsi
(8)
Stress Analysis.
First, determine the reaction forces at bearings A and B. The free body diagram of the shaft is shown
in Fig. 1a.
11
Figure 1a. The free body diagram of the shaft.
The sum of the moments about the right-hand bearing B can be written as
M
B
0
(9a)
that is
24 RA  16 F1  8 F2  0
(9b)
Therefore, the reaction force at the left-hand bearing A is
RA 
2500(16)  1000(8)
 2000 lb
24
(9c)
The sum of the forces in the Y-direction can be written as
F
y
0
(10a)
that is
RA  RB  F1  F2  0
(10b)
2000  RB  2500  1000  0
(10c)
or
Therefore, the reaction force at bearing B is
RB  1500 lb
(10c)
The shear force between the left-hand bearing A and the force F1 is
VA  RA   2000 lb
(11a)
The shear force between the forces F1 and F2 is
V  RA  F1   2000  2500   500 lb
(11b)
The shear force between the force F2 and the right-hand bearing B is
VB   RB   1500 lb
12
(11c)
The shear force diagram of the shaft is shown in Fig. 1b
Figure 1b. The shear force diagram.
Let the point of application of the force F1 be denoted as point C and the point of application of the
force F2 be denoted as point D. Therefore, the bending moment at point C is
M C  8 VA  8 x 2000  16000 lb.in
(12a)
Also, the bending moment at point D is
M D  16VA  8 F1  16 x 2000  8 x 2500  12000 lb.in
(12b)
The bending-moment diagram of the shaft is shown in Fig. 1c.
Figure 1c. The bending moment digram.
The conclusion from Figs. 1a, 1b, and 1c is that the critical section of the shaft is at the shoulder
fillet between the 1-5/8 in and the 1-7/8 in diameters where: (i) the bending moment is large, (ii) the
diameter is smaller, and (iii) stress concentration exists.
Note that the fully reversed bending moment at the shoulder fillet is
M  16000  2.5 x 500  14750 lb.in
(13)
The critical element is on the outer surface of the shaft at this critical section. The fatigue stress
concentration factor can be written from Eq. (6-32), see page 295, as
K f  1  q( K t  1)
First, determine the theoretical stress concentration factor. The known data for the shaft is
13
(14)
D  1.875 in,
d  1.625 in
and
r  0.0625 in
(15a)
Therefore, the geometric ratios are
r 0.0625

 0.04
d 1.625
and
D 1.875

 1.15
d 1.625
(15b)
Therefore, the theoretical stress concentration factor from Figure A-15-9, see page 1028, is
K t  1.95
(16)
The notch sensitivity can be obtained from either Fig. 6-20, or from the curve-fit, see Eqs. (6-34)
and (6-35a), page 296. Here we will use the equations. Therefore, the notch sensitivity can be written
from Eq. (6-34), page 296, as
1
(17)
q
a
1
r
The Neuber constant from Eq. (6-35a), page 296, is
a  0.246  (3.08  103 ) Sut  (1.51 105 ) Sut 2  (2.67 108 ) Sut 3 in
(18a)
Note that the ultimate tensile strength in Eq. (17a) must be in kpsi. Substituting Eq. (2b) into Eq. (18a),
the Neuber constant is
a  0.246  (3.08 103 )85  (1.51 105 )852  (2.67  108 )853  0.07690 in
(18b)
Substituting Eq. (18b) into Eq. (17), the notch sensitivity is
q
1
 0.765
0.07690
1
0.0625
(19)
Substituting Eqs. (16) and (19) into Eq. (14), the fatigue stress concentration factor is
K f  1  0.765  (1.95  1)  1.727
(20)
The normal stress due to the fully reversed bending moment on the critical element (this is the zero
mean stress problem) can be written as
Mc
 rev 
(21a)
I
Substituting Eq. (13) and the known geometry into Eq. (21a), the normal stress is
 rev 
14750  (1.625 / 2)
 35.0 kpsi
( / 64) 1.6254
14
(21b)
Since this problem is a zero mean stress problem then the factor of safety guarding against fatigue
failure can be written as
S
Se
(22a)
Nf  e 
 a K f  rev
Substituting Eqs. (8), (20), and (21b) into Eq. (22a), the fatigue factor of safety is
Nf 
29.526
 0.488
1.727  35.0
(22b)
Since the fatigue factor of safety is less than one then the shaft has a finite life. The S  N diagram can
be used to determine the life of the shaft. The number of cycles to failure can be written from Eq. (6-16)
see page 285, as
1/ b
1/ b
 K f  rev 
a 
(23)
N   

 a 
 a 
The constant can be written from Eq. (6-14), see page 285, as
a
( f Sut ) 2
Se
(24)
The fatigue strength fraction from Fig. 6-18, see page 285, is
f  0.867
(25)
Substituting Eqs. (2) and (25) into Eq. (24), the coefficient is
a
(0.867  85) 2
 183.938 kpsi
29.526
(26)
The exponent can be written from Eq. (6-15), see page 285, as
 f Sut 
1
b   log 

3
 Se 
(27)
Substituting Eqs. (2), (8), and (26) into Eq. (27), the exponent is
1
 0.876  85 
b   log 
   0.134
3
 29.526 
(28)
Substituting Eqs. (19), (21b), (26), and (28) into Eq. (23), the number of cycles to failure is
1
 1.727  35.0  0.134
N 
 4069.047 cycles

 183.938 
Therefore, the number of cycles to failure can be taken as
15
(29a)
N  4.07 103 cycles
(29b)
Check for yielding. The factor of safety guarding against yielding (see Chapter 5 or use the Langer line)
for the critical element on the outer surface of the shaft at the critical section can be written as
Ny 
Sy
a

Sy
 rev
(30)
The normal stress due to the fully reversed bending moment on the critical element is given by Eq.
(21a), that is
Mc
 rev 
(31)
I
Since the fatigue stress concentration factor will not be included in this equation (see Section 3. 13, page
110) then the maximum bending moment, see Fig. 1b, is taken as
M  16000 lb.in
(32)
Substituting Eq. (32) and the known geometry into Eq. (31), the normal stress is
 rev 
16000  (1.625 / 2)
 37.98 kpsi
( / 64) 1.6254
(33)
The yield strength of AISI CD steel, see Table A-20, page 1040, is
S y  71 kpsi
(34)
Substituting Eqs. (33) and (34) into Eq. (30), the factor of safety guarding against yielding for this
element is
71
(35)
Ny 
 1.87
37.98
Note that the completely reversed bending stress is much less than the yield strength of the shaft
material, therefore, yielding is not predicted to have occurred.
Aside: Consider the critical element that was investigated in the fatigue case. The maximum bending
moment acting at this critical section is
M  14750 lb.in
(36)
Substituting Eq. (36) and the known geometry into Eq. (31), the normal stress is
 rev 
14750  (1.625 / 2)
 35.0 kpsi
( / 64) 1.6254
(37)
Substituting Eqs. (34) and (37) into Eq. (30), the factor of safety guarding against yielding for the
critical element is
71
Ny 
 2.01
(38)
35.0
16