Transcript Fatigue_example_3 - The George W. Woodruff School of

ME3180
ME 3180B - Mechanical Engineering Design - Spring 2005
Example #3
(example 7-26 in text)
The George W. Woodruff School of Mechanical Engineering
ME3180
Example #3 : Problem 6-23 Shigley
Given:
• The figure shows the free-body diagram of a connecting-link portion
having stress concentration at three places. The dimensions are: r = 0.25in,
d = 0.75in, h = 0.50in, w1=3.75in, and w2 = 2.5in. The forces F fluctuate
between a tension of 4 Kip and a compression of 16 Kip.
• Neglect column action. Find the least factor of safety if material is colddrawn AISI 1018 steel.
The George W. Woodruff School of Mechanical Engineering
ME3180
Example #3: Problem 6-23 Shigley
Solution:
• From Table A-20 (Shiggley)
– Sut = 64 Kpsi
– Syt = 54 Kpsi
• From your note:
– S’e = 0.504 Sut = 32.3Kpsi
– (Note: You could use S’e =0.45Sut in all your calculations. If you do,
then set CL = 1 and you will still obtain the same result)
• Table 7-4 in your note:
– a = 2.7Kpsi, b = -0.265
–
–
–
–
CF = Ka = aSbut = 2.7 Kpsi (64Kpsi)-0.265
CF = Ka = 0.897
Cs = Kb =1, CL = Kc = 0.923 (Equation 7-22)
Se = CFCsCILCRCTKeS’e (Equation 7-13)
= 0.897(1)(0.923)(32.3) Kpsi (The value for Ke is not included here,
you can include it if you want to)
= 26.7 kpsi
The George W. Woodruff School of Mechanical Engineering
ME3180
Example #3 : Problem 6-23 Shigley
Solution: See Figure A-15-5 Shigley
• Check for yielding at the fillet
F
 c  min 
16
 12.8Kpsi
(2.5)(0.5)
W2 h
S yc
 S yt
 54
nstatic 


 4.22
c
c
 12.8
fig. A  15  5
Norton: Fig E-9 (page 998)
• If you use 4 Kip, n will be > 4.22.
– 4 Kpsi/(2.5*0.5) = 3.2Kpsi
– n = 54/3.2 = 16.9
The George W. Woodruff School of Mechanical Engineering
ME3180
Example #3 : Problem 6-23 Shigley
Solution:
• Check for fatigue Failure ( Calculate Kt)
D  3.75, d  2.5
w1
D 3.75


 1.5
2
.
5
w2
d
r
0.25

 0.1
d
2.5
 K t  2.1
• Equation 5-26:
– Kf = 1 + q(Kt-1)
• Figure 9-2: (use r = 0.25in)
– q = 0.78 (when r ≥0.16”, use the value at r = 0.16”)
The George W. Woodruff School of Mechanical Engineering
ME3180
Example #3 : Problem 6-23 Shigley
Solution:
K f  1  0.78(2.1  1)  1.86
4kip
4kip
 max 

 3.2kpsi
2
2
(2.5)(0.5)in
1.25in
 min 
 16kip
 12.8kpsi
2
1.25in
a  K f
 max   min
(useequation7  31)
2
3.2  (12.8)
 a  1.86
2
 a  14.88kpsi
S
26.7
n e 
 1.8
 a 14.88
m 
 max   min 3.2  (12.8)

kpsi  4.8kpsi
2
2
The George W. Woodruff School of Mechanical Engineering
ME3180
Example #3 : Problem 6-23 Shigley
Solution:
• Check for failure due to Soderberg, Goodman and Gerber failure criteria.
– Soderberg
a
m
1


Se
S yt
n
14.88 4.8
1


26.7
54
n
 n  2.13
The George W. Woodruff School of Mechanical Engineering
ME3180
Example #3 : Problem 6-23 Shigley
Solution:
– Goodman
a
m
1


S e Sut n
14.88 4.8 1


26.7
64 n
n  2.07
The George W. Woodruff School of Mechanical Engineering
ME3180
Example #3 : Problem 6-23 Shigley
Solution:
– Gerber
2

 1


2
14.88n  4.8 
 
n  1
26.7
 64 
0.0056n 2  0.5573n  1  0
n a  n m
 
Se
 Sut
n  2.07
The George W. Woodruff School of Mechanical Engineering
ME3180
Example #3 : Problem 6-23 Shigley
Solution: See Figure A-15-1 Page 1006 Shigley
• Check for yielding at hole
4kip
4kip
4kip


h( w1  d ) 0.5(3.75  0.75)in2 0.5(3)in2
 2.6 kpsi
 t   max 
n
S yt
 max

54
 20.26
2. 6
 c   min 
n
 S yt
 min

 16kip
 10.67kpsi
0.5(3)in2
 54
 5.06
 10.67
The George W. Woodruff School of Mechanical Engineering
ME3180
Example #3 : Problem 6-23 Shigley
Solution:
• Check for fatigue failure (Calculate Kt )
d
0.75

 0.20
w1 3.75
K t  2.5
use 0.20 in Fig. A-15-1 or Fig (E-13) to obtain Kt
• q = 0.78 (use r = 0.375’’ in fig. 5-16) Kf = 1+ 0.78(2.5-1) = 2.17
a 
 max   min
2
2.67  (10.67)
 a  2.17
 14.47kpsi
2
S
26.7
n e 
 1.85
 a 14.47
The George W. Woodruff School of Mechanical Engineering
ME3180
Example #3 : Problem 6-23 Shigley
Solution:
• Let us see what we get with Soderberg Relation
 max   min 3.2  12.8
m 

 4.8kpsi
2
a m 1


S e S yt n
14.47  4 1


26.7 54 n
n  2.14
2
• You can do the same for the other two-Gerber and Goodman.
• Failure will occur at the discontinuity with the smallest factor of safety
The George W. Woodruff School of Mechanical Engineering
ME3180
Example #3
The George W. Woodruff School of Mechanical Engineering
ME3180
Example #3
The George W. Woodruff School of Mechanical Engineering
ME3180
Example #3
The George W. Woodruff School of Mechanical Engineering
ME3180
Example #3
The George W. Woodruff School of Mechanical Engineering