102
Chapter 2
levels obtained. The material parameters will be determined for ~e first.
4340 steel 1/2-in diameter aut = 210 kpsi is heat treated and drawn to 800°F.
O’e = kakbkc.., k l ate
(2.106)
a’ e Fig. 2.26.
-t Gut
Gut Z e __ a,t 2 __ 4-0.0667 =
O" e = ~- ~e -- 30 O" e-’ 30
aut
ka Eq. (2.57) surface finish and Table 2.4
~a = 0.947 - 0.159 x 10-saut = 0.6131 with ~a = 4-0.0406 and
~ 0.0406
Cv - ka - 0.613~ - 0.066
kb (E_q. (2.60)) size shape_d<2"
k6=0.85
~b=0.06kb
C~=0.06
kc Section A.3 Reliability coupling Eq. (2.42)
kc = 1
C~ = 0
kd Section A.4 Temperature ka = 1 since oil operates @ 180°F C~ = 0
ke Section A.5 Apply stress contrations on stresses ke = 1
Cv = 0
kf Eq. 2.72 Shot peening ~f = [(1.22 - 1.13)/3] = 4-0.03, ~:f = 1.13,
Cv = 0.0265
kg Section A.7 Internal structure
kh Section A.8 Environment (in oil)
ki Section A.9 No surface treatment
kj Section A.10 Fretting
kk Section A. 11 Shock in stress calculation
k~ Section A.12 Radiation
km Eq. 2.90 10 7 cycles
Evaluation per Example 2.18
5"e = (0.6131)(0.85)(1.13)(0.95) 000 psi)
6"~= 58,741 psi
~
-t
C
2
C
2
za~ = a e w’~ 4v~
l=1
= 8-e[(0.0667)2+) (0.066) 2 + (0.06) 2 + (0.0265)2]
~/2
Z~ e -- = 4-0.1145
~aut = ~0.05 (Eq. (2.54))
~ut
kg= l
C,,=O
kh= 1
Cv=O
ki---- 1
Cv=O
k~=0.95
C~=0
kk= 1
C~=0
kt = 1
C~ = 0
km ~- 1
C~ = 0
(2.107)